Chapter 4 Principle of Mathematical Induction (Additional Questions)

A mathematical statement (also known as a proposition) is a declarative sentence that is either true or false, but not both.

Let's analyze each option based on this definition:

(A) "The sum of all prime numbers is 100."

This is a declarative sentence. It makes a claim about the sum of all prime numbers. The sum of all prime numbers ($2, 3, 5, 7, ...$) is not $100$; it is generally considered to be infinite. Therefore, this statement is false. Since it is a sentence that is either true or false (specifically, false), it is a mathematical statement.

(B) "Tomorrow is Sunday."

This is a declarative sentence, so it is a statement. However, its truth value depends on the day it is spoken. It is not a statement about mathematical objects or properties within a mathematical framework. Therefore, it is not considered a mathematical statement in this context.

(C) "The product of two even integers is an even integer."

This is a declarative sentence. It makes a claim about a property of even integers under multiplication. This statement is always true. We can prove it as follows: Let $a$ and $b$ be any two even integers. By definition, an even integer is an integer that can be expressed in the form $2k$, where $k$ is an integer. So, we can write $a = 2k_1$ and $b = 2k_2$ for some integers $k_1$ and $k_2$. Their product is $ab = (2k_1)(2k_2) = 4k_1k_2 = 2(2k_1k_2)$. Since $k_1$ and $k_2$ are integers, $2k_1k_2$ is also an integer. Therefore, $ab$ is of the form $2 \times (\text{an integer})$, which means $ab$ is an even integer.

Since the statement is always true, it has a fixed truth value and is a statement about a fundamental mathematical property. Therefore, it is a mathematical statement.

(D) "Open the door."

This is an imperative sentence (a command). It is not a declarative sentence and cannot be assigned a truth value (it is neither true nor false). Therefore, it is not a statement.

Both options (A) and (C) fit the definition of a mathematical statement as they are declarative sentences that are either true or false. However, option (C) describes a general, provable property of integers, which is a classic example used to illustrate mathematical statements and theorems in logic and mathematics. While (A) is also a mathematical statement (a false claim about prime numbers), (C) is likely the intended best answer in a multiple-choice question designed to identify fundamental mathematical propositions.

The correct option is (C) The product of two even integers is an even integer.

The Principle of Mathematical Induction is a fundamental method used to prove that a statement $P(n)$ is true for every positive integer $n$ (or for every integer $n$ greater than or equal to some fixed integer).

The principle consists of two main steps:

1. Base Case (or Basis Step): Prove that the statement $P(n)$ is true for the initial value of $n$. This initial value is typically $n=1$ when proving statements for all positive integers, or $n=0$ when proving statements for all non-negative integers, or some other integer depending on the set of integers for which the statement is claimed to be true.

2. Inductive Step (or Inductive Hypothesis): Assume that the statement $P(k)$ is true for some arbitrary integer $k$ (where $k$ is greater than or equal to the initial value from the base case). This assumption is called the inductive hypothesis.

Then, prove that the statement $P(k+1)$ is true, using the assumption that $P(k)$ is true.

Let's examine the given options in the context of the Base Case:

(A) $n=0$: This is a possible initial value for the base case if the statement is about non-negative integers ($0, 1, 2, ...$). However, it's not the standard base case when discussing "positive integers".

(B) $n=1$: This is the standard initial value for the base case when the statement is about positive integers ($1, 2, 3, ...$).

(C) Some positive integer $k$: This is the integer used in the Inductive Step (the inductive hypothesis assumes $P(k)$ is true), not the specific value used in the Base Case.

(D) All positive integers $n$: This is the set of values for which the statement is proven to be true after successfully completing both the Base Case and the Inductive Step. It is the conclusion of the induction, not the Base Case itself.

In the most common formulation of the Principle of Mathematical Induction for positive integers, the first step (Base Case) is to prove the statement for $n=1$.

The correct option is (B) $n=1$.

The Principle of Mathematical Induction is used to prove statements about integers.

It consists of two main steps:

1. Base Case: Prove the statement $P(n)$ for the initial value of $n$ (e.g., $n=1$ for positive integers).

2. Inductive Step: This step establishes that if the statement holds for some arbitrary integer $k$ (the inductive hypothesis), it must also hold for the next integer, $k+1$.

Specifically, the Inductive Step involves the following:

Assume that $P(k)$ is true for some positive integer $k$ (or $k \geq$ the initial value from the Base Case). This is called the Inductive Hypothesis.

Using this assumption, prove that the statement $P(k+1)$ is also true.

Let's look at the options provided for what needs to be proven in the Inductive Step:

(A) $P(k-1)$ is true: Proving the statement for a previous value is not the goal of the standard inductive step.

(B) $P(k+1)$ is true: This is precisely what needs to be proven in the inductive step, assuming $P(k)$ is true.

(C) $P(n)$ is true for all $n > k$: This is the overall goal for all integers greater than $k$ (starting from $k+1$), which is achieved by successfully completing the inductive step for an arbitrary $k$, not what is proven in a single step.

(D) $P(n)$ is true for all $n \leq k$: Proving the statement for values up to $k$ is either part of the premise (for $n=k$) or not what the inductive step aims to show.

Therefore, the Inductive Step involves assuming $P(k)$ is true and proving that $P(k+1)$ is true.

The correct option is (B) $P(k+1)$ is true.

The Principle of Mathematical Induction (PMI) is a technique used to prove that a statement $P(n)$ holds for all integers $n$ in a specified range (usually starting from a specific integer). It is particularly suited for statements about positive integers or non-negative integers.

The statement given is $P(n): 1 + 2 + 3 + ... + n = \frac{n(n+1)}{2}$.

The left side of the equation represents the sum of the first $n$ positive integers. This sum is well-defined only when $n$ is a positive integer, i.e., $n \geq 1$. For example, if $n=3.5$ or $n=-2$, the concept of "the sum of the first $n$ positive integers" doesn't typically apply in this context.

The Principle of Mathematical Induction is applied as follows to prove this statement:

1. Base Case: Prove $P(1)$ is true.

Left side: $1$. Right side: $\frac{1(1+1)}{2} = \frac{1 \times 2}{2} = 1$. Since $1=1$, $P(1)$ is true.

2. Inductive Step: Assume $P(k)$ is true for some positive integer $k \geq 1$. That is, assume $1 + 2 + 3 + ... + k = \frac{k(k+1)}{2}$. Prove $P(k+1)$ is true, i.e., prove $1 + 2 + 3 + ... + k + (k+1) = \frac{(k+1)((k+1)+1)}{2} = \frac{(k+1)(k+2)}{2}$.

Starting with the left side of $P(k+1)$:

$1 + 2 + 3 + ... + k + (k+1)$

By the inductive hypothesis, $1 + 2 + 3 + ... + k = \frac{k(k+1)}{2}$.

So, $1 + 2 + 3 + ... + k + (k+1) = \frac{k(k+1)}{2} + (k+1)$

Common denominator:

$= \frac{k(k+1) + 2(k+1)}{2}$

Factor out $(k+1)$:

$= \frac{(k+1)(k+2)}{2}$

This is the right side of $P(k+1)$. Thus, if $P(k)$ is true, then $P(k+1)$ is true.

By the Principle of Mathematical Induction, since $P(1)$ is true and $P(k) \implies P(k+1)$ for all positive integers $k$, the statement $P(n)$ is true for all positive integers $n$. Positive integers are integers $n \geq 1$.

Looking at the options:

(A) All real numbers $n$: PMI is not used for all real numbers.

(B) All integers $n \geq 1$: This is the set of positive integers, which is the domain for which the formula is proven true by induction.

(C) All rational numbers $n$: PMI is not used for all rational numbers.

(D) $n=1$ only: The base case proves it for $n=1$, but the induction proves it for *all* integers $n \geq 1$.

The correct option is (B) All integers $n \geq 1$.

The given statement is $P(n): 2^n > n$.

In the Principle of Mathematical Induction, the base case involves verifying the statement $P(n)$ for the initial value of $n$. The question specifies the base case is $n=1$.

To find what $P(1)$ states, we substitute $n=1$ into the statement $P(n): 2^n > n$.

$P(1)$ is the statement $2^1 > 1$.

Evaluating this statement:

$2^1 = 2$

So, $P(1)$ is $2 > 1$.

Let's check the options:

(A) $2^0 > 0$, which is $1 > 0$ (True): This is the statement $P(0)$, not $P(1)$.

(B) $2^1 > 1$, which is $2 > 1$ (True): This is the statement $P(1)$, and it is true.

(C) $2^2 > 2$, which is $4 > 2$ (True): This is the statement $P(2)$, not $P(1)$.

(D) $2^n > n$ for some $n=1$ (True): This phrase describes the evaluation of $P(n)$ at $n=1$, but the statement itself is $2^1 > 1$. Option (B) directly provides the statement $P(1)$.

The statement $P(1)$ is $2^1 > 1$, which simplifies to $2 > 1$. This statement is true.

The correct option is (B) $2^1 > 1$, which is $2 > 1$ (True).

The given statement is $P(n): 1 + 3 + 5 + ... + (2n-1) = n^2$. This statement represents the sum of the first $n$ odd positive integers.

The statement $P(k)$ is assumed to be true, which is:

To prove $P(k+1)$ is true, we need to show the statement holds for $n = k+1$. The left side of $P(k+1)$ is the sum of the first $k+1$ odd positive integers. The $n$-th odd positive integer is given by the formula $2n-1$. So, the $(k+1)$-th odd positive integer is obtained by replacing $n$ with $k+1$ in the formula $2n-1$:

$(k+1)$-th term = $2(k+1)-1$

$(k+1)$-th term = $2k + 2 - 1$

$(k+1)$-th term = $2k + 1$

The statement $P(k+1)$ is:

1 + 3 + 5 + ... + $(2k-1)$ + $(2(k+1)-1)$ = $(k+1)^2$

Which simplifies to:

1 + 3 + 5 + ... + $(2k-1)$ + $(2k+1)$ = $(k+1)^2$

Comparing the left side of $P(k+1)$ with the left side of $P(k)$ (equation (i)), we see that the terms in $P(k+1)$ are the terms in $P(k)$ plus one additional term. This additional term is the $(k+1)$-th term in the series, which we calculated as $2k+1$.

Left side of $P(k+1) = \underbrace{1 + 3 + 5 + ... + (2k-1)}_{ \text{Left side of } P(k) } + (2k+1)$

Therefore, the term added to the left side when going from $P(k)$ to $P(k+1)$ is $2k+1$.

Let's check the options:

(A) $2k-1$: This is the $k$-th term of the series, not the term added to get to the $(k+1)$-th sum.

(B) $2k+1$: This is the $(k+1)$-th term of the series, which is added to the sum of the first $k$ terms to get the sum of the first $k+1$ terms.

(C) $(k+1)^2$: This is the right side of the statement $P(k+1)$, not a term added to the left side sum.

(D) $k^2$: This is the right side of the statement $P(k)$ (the assumed sum of the first $k$ terms), not a term added to the left side sum.

The correct option is (B) $2k+1$.

The statement to be proven by PMI is $P(n): n^2 + n$ is an even integer.

We need to prove this for all positive integers $n$. The set of positive integers starts with $1, 2, 3, ...$.

The first step in the Principle of Mathematical Induction is the Base Case. We need to show that the statement $P(n)$ is true for the smallest value of $n$ in the domain, which is $n=1$ for positive integers.

To find the base case $P(1)$, we substitute $n=1$ into the statement $P(n): n^2 + n$ is an even integer.

$P(1)$ is the statement: $1^2 + 1$ is an even integer.

Let's evaluate $1^2 + 1$:

$1^2 + 1 = 1 + 1 = 2$

So, the statement $P(1)$ is "$2$ is an even integer". This statement is true, as $2 = 2 \times 1$, where $1$ is an integer.

Let's examine the options:

(A) $1^2 + 1 = 2$, which is even (True): This accurately states $P(1)$ and its truth value.

(B) $2^2 + 2 = 6$, which is even (True): This is the statement $P(2)$, not $P(1)$.

(C) $n^2 + n$ is even for $n=1$ (True): This is a description of the base case, but option (A) provides the specific calculation and result for $n=1$.

(D) $1^2 + 1 = 2$: This is the calculation for $n=1$, but it does not state that the result (2) is even, which is the property being checked for $P(n)$.

The most complete and correct description of the base case $P(1)$ among the options is the one that shows the calculation for $n=1$ and verifies the property (being even).

The correct option is (A) $1^2 + 1 = 2$, which is even (True).

The statement to be proven by the Principle of Mathematical Induction (PMI) is $P(n): n < 2^n$.

The question asks for the correct base case to start the induction for all positive integers $n$.

The set of positive integers is $\{1, 2, 3, 4, ...\}$.

The Principle of Mathematical Induction for a statement $P(n)$ on positive integers starts with verifying the base case for the smallest positive integer, which is $n=1$.

So, the base case is $P(1)$.

Let's check what $P(1)$ means by substituting $n=1$ into the statement $P(n): n < 2^n$:

$P(1)$ is the statement $1 < 2^1$.

Evaluating $2^1$, we get $2$. So, $P(1)$ is the statement $1 < 2$. This is a true statement.

Let's consider the given options:

(A) $P(0)$: This would be the base case if the statement was about non-negative integers ($n \geq 0$). $P(0)$ is $0 < 2^0$, i.e., $0 < 1$, which is true. But the question is about positive integers.

(B) $P(1)$: This is the base case for positive integers. $P(1)$ is $1 < 2^1$, i.e., $1 < 2$, which is true.

(C) $P(2)$: This could be a base case if the statement was about integers $n \geq 2$. $P(2)$ is $2 < 2^2$, i.e., $2 < 4$, which is true.

(D) $P(3)$: This could be a base case if the statement was about integers $n \geq 3$. $P(3)$ is $3 < 2^3$, i.e., $3 < 8$, which is true.

Since we are asked for the base case for "all positive integers", we must start with the smallest positive integer, which is $1$.

The correct base case is $P(1)$.

The correct option is (B) $P(1)$.

The statement is $P(n): 7^n - 3^n$ is divisible by 4 for all positive integers $n$.

We are in the Inductive Step. We assume $P(k)$ is true for some positive integer $k$.

Inductive Hypothesis: $P(k)$ is true, meaning $7^k - 3^k$ is divisible by 4. This can be written as:

We need to prove $P(k+1)$ is true, meaning $7^{k+1} - 3^{k+1}$ is divisible by 4.

We start with the expression $7^{k+1} - 3^{k+1}$ and try to use the inductive hypothesis ($7^k - 3^k = 4m$).

Let's manipulate the expression $7^{k+1} - 3^{k+1}$:

$7^{k+1} - 3^{k+1} = 7 \cdot 7^k - 3 \cdot 3^k$

We want to introduce the term $7^k - 3^k$. A common technique is to add and subtract a term involving $7^k$ or $3^k$ multiplied by one of the coefficients (7 or 3). Let's try multiplying the inductive hypothesis term $(7^k - 3^k)$ by 7:

$7(7^k - 3^k) = 7 \cdot 7^k - 7 \cdot 3^k$

Now consider $7^{k+1} - 3^{k+1} = 7 \cdot 7^k - 3 \cdot 3^k$. We can rewrite $7 \cdot 7^k$ as $7(7^k - 3^k + 3^k) = 7(7^k - 3^k) + 7 \cdot 3^k$.

So, $7^{k+1} - 3^{k+1} = (7(7^k - 3^k) + 7 \cdot 3^k) - 3 \cdot 3^k$

$= 7(7^k - 3^k) + 7 \cdot 3^k - 3 \cdot 3^k$

$= 7(7^k - 3^k) + (7 - 3) 3^k$

$= 7(7^k - 3^k) + 4 \cdot 3^k$

Using the inductive hypothesis $7^k - 3^k = 4m$, we can substitute this into the expression:

$7^{k+1} - 3^{k+1} = 7(4m) + 4 \cdot 3^k$

$= 4(7m) + 4 \cdot 3^k$

$= 4(7m + 3^k)$

Since $m$ and $k$ are integers, $7m + 3^k$ is an integer. Therefore, $7^{k+1} - 3^{k+1}$ is divisible by 4.

Now let's compare our rewritten expression with the given options:

(A) $7(7^k - 3^k) + 4 \cdot 3^k$: This matches our rewritten expression.

(B) $7(7^k) - 3(3^k)$: This is the original expanded form of $7^{k+1} - 3^{k+1}$, not a rewritten form useful for the inductive step.

(C) $7(4m) + 3^k(7-3)$: This uses the inductive hypothesis substitution $7^k - 3^k = 4m$ prematurely within a form that isn't one of the options.

(D) $7(7^k - 3^k) + 3^k(7-3)$: This also matches our rewritten expression, as $7-3=4$. This is the same as option (A).

Both (A) and (D) represent the correct rewriting. Option (A) is $7(7^k - 3^k) + 4 \cdot 3^k$ and Option (D) is $7(7^k - 3^k) + 3^k(7-3)$, which is $7(7^k - 3^k) + 4 \cdot 3^k$. Since they are identical, and (A) is listed, it is the correct choice.

The correct option is (A) $7(7^k - 3^k) + 4 \cdot 3^k$.

The statement is $P(n): 10^n + 3 \cdot 4^{n+2} + 5$ is divisible by 9 for all positive integers $n$.

The first step in the Principle of Mathematical Induction is the Base Case. We need to show that the statement $P(n)$ is true for the smallest positive integer, which is $n=1$.

To verify the base case $P(1)$, we substitute $n=1$ into the expression $10^n + 3 \cdot 4^{n+2} + 5$ and check if the result is divisible by 9.

Substitute $n=1$:

$10^1 + 3 \cdot 4^{1+2} + 5$

$= 10^1 + 3 \cdot 4^3 + 5$

Calculate the values:

$10^1 = 10$

$4^3 = 4 \times 4 \times 4 = 16 \times 4 = 64$

Substitute back into the expression:

$= 10 + 3 \cdot 64 + 5$

$= 10 + 192 + 5$

$= 202 + 5$

$= 207$

So, for $n=1$, the expression evaluates to 207. The base case requires verifying if this value, 207, is divisible by 9.

To check for divisibility by 9, we can sum the digits of 207: $2 + 0 + 7 = 9$. Since the sum of the digits is 9 (which is divisible by 9), the number 207 is divisible by 9.

$207 \div 9 = 23$. So, $207 = 9 \times 23$.

The base case $P(1)$ is verified by calculating the expression for $n=1$ and confirming that the result is divisible by 9.

Let's examine the options:

(A) $10^1 + 3 \cdot 4^{1+2} + 5 = 10 + 3 \cdot 4^3 + 5 = 10 + 3 \cdot 64 + 5 = 10 + 192 + 5 = 207$. Is 207 divisible by 9?: This option describes the calculation and the question that needs to be answered for the base case.

(B) $10^1 + 3 \cdot 4^{1+2} + 5 = 207$. 207 = 9 $\times$ 23. So it is divisible by 9.: This option performs the calculation and confirms the divisibility by 9, concluding that the base case holds.

(C) The statement is true for $n=1$ if $207$ is divisible by 9.: This option correctly states the condition for the base case to be true.

(D) All of the above steps are part of verifying the base case for $n=1$.: Verifying the base case for $n=1$ involves calculating the expression, checking if the result (207) is divisible by 9, and concluding that the statement is true for $n=1$. Options (A), (B), and (C) all describe aspects of this verification process. Option (A) shows the calculation and the question, Option (B) shows the calculation and the answer (divisibility), and Option (C) states the condition for the base case's truth. Collectively, these steps constitute the verification of the base case. Therefore, all the steps mentioned in (A), (B), and (C) are part of the process.

The correct option is (D) All of the above steps are part of verifying the base case for $n=1$.

Let's analyze the Assertion (A) and the Reason (R).

Assertion (A): The statement $P(n): n! > 2^n$ is true for all positive integers $n \geq 4$.

This is a statement about an inequality involving factorials and powers of 2. We can test a few values of $n \geq 4$:

• For $n=4$: $4! = 4 \times 3 \times 2 \times 1 = 24$. $2^4 = 2 \times 2 \times 2 \times 2 = 16$. $24 > 16$, so $P(4)$ is true.
• For $n=5$: $5! = 5 \times 4! = 5 \times 24 = 120$. $2^5 = 2 \times 2^4 = 2 \times 16 = 32$. $120 > 32$, so $P(5)$ is true.
• For $n=6$: $6! = 6 \times 5! = 6 \times 120 = 720$. $2^6 = 2 \times 2^5 = 2 \times 32 = 64$. $720 > 64$, so $P(6)$ is true.

The assertion claims this holds for all $n \geq 4$. This can be proven using Mathematical Induction with the base case $n=4$. So, Assertion (A) appears to be true.

Reason (R): This provides a potential proof using Mathematical Induction.

1. Base Case: For $n=4$, $4! = 24$ and $2^4 = 16$, so $24 > 16$, which is true. This correctly verifies the base case for $n=4$.

2. Inductive Step: Assume $P(k)$ is true for some integer $k \geq 4$. That is, assume $k! > 2^k$.

We need to prove $P(k+1)$ is true, i.e., $(k+1)! > 2^{k+1}$.

Start with $(k+1)!$:

$(k+1)! = (k+1) \cdot k!$

By the inductive hypothesis, $k! > 2^k$. So, multiplying both sides by $(k+1)$ (which is positive since $k \geq 4$), the inequality holds:

$(k+1) \cdot k! > (k+1) \cdot 2^k$

So, $(k+1)! > (k+1) \cdot 2^k$.

Now we need to show that $(k+1) \cdot 2^k > 2^{k+1}$.

$2^{k+1} = 2 \cdot 2^k$.

The inequality we need to show is $(k+1) \cdot 2^k > 2 \cdot 2^k$.

We can divide both sides by $2^k$ (which is positive):

$k+1 > 2$

This inequality $k+1 > 2$ is equivalent to $k > 1$.

The Reason states "which is true for $k \geq 2$". This is correct. However, the base case started at $n=4$, so our inductive hypothesis applies for $k \geq 4$. If the inductive hypothesis holds for $k \geq 4$, then $k > 1$ is certainly true.

The Reason then concludes: "Since the base case starts at $n=4$, the induction holds."

The argument in the Reason correctly demonstrates that if $P(k)$ is true for $k \geq 4$, then $P(k+1)$ is true. Combined with the verified base case $P(4)$, this proves that $P(n)$ is true for all integers $n \geq 4$ by PMI.

Thus, the Reason (R) provides a correct proof by induction for the statement in Assertion (A), starting from $n=4$. It correctly explains why the statement is true for $n \geq 4$ by showing the base case holds and the inductive step works for the relevant values of $k$.

Both Assertion (A) and Reason (R) are true, and Reason (R) provides a correct explanation and justification for Assertion (A) using the Principle of Mathematical Induction.

The correct option is (A) Both A and R are true and R is the correct explanation of A.

The Principle of Mathematical Induction (PMI) is a powerful proof technique used to prove statements that are asserted to be true for all natural numbers or all integers greater than or equal to a specific integer (the base case).

A statement that can be proven by PMI typically takes the form $P(n)$, where $P$ is some property and $n$ is an integer variable, usually a positive integer.

Let's examine each option:

(A) "The sun rises in the east."

This is a statement about a natural phenomenon. While it is generally considered true, it is not a mathematical statement depending on an integer variable $n$. It cannot be proven using PMI.

(B) "The equation $x^2 = 4$ has two solutions."

This is a mathematical statement about a specific equation. It is true (the solutions are $x=2$ and $x=-2$), but it does not involve a property that varies with a positive integer $n$ in a way suitable for induction. It cannot be proven using PMI.

(C) "For every positive integer $n$, $1 \cdot 2 + 2 \cdot 3 + ... + n(n+1) = \frac{n(n+1)(n+2)}{3}$."

This is a statement about a summation formula that depends on a positive integer $n$. This is a typical form of statement that can be proven using the Principle of Mathematical Induction. Let $P(n)$ be the statement $1 \cdot 2 + 2 \cdot 3 + ... + n(n+1) = \frac{n(n+1)(n+2)}{3}$.

To prove this by PMI, we would:

Base Case: Verify $P(1)$. Left side: $1(1+1) = 1 \cdot 2 = 2$. Right side: $\frac{1(1+1)(1+2)}{3} = \frac{1 \cdot 2 \cdot 3}{3} = \frac{6}{3} = 2$. Since $2=2$, $P(1)$ is true.

Inductive Step: Assume $P(k)$ is true for some positive integer $k$, i.e., $1 \cdot 2 + ... + k(k+1) = \frac{k(k+1)(k+2)}{3}$. We then need to prove $P(k+1)$ is true, i.e., $1 \cdot 2 + ... + k(k+1) + (k+1)((k+1)+1) = \frac{(k+1)((k+1)+1)((k+1)+2)}{3}$. The process involves using the assumption $P(k)$ to show that $P(k+1)$ follows. This is feasible for this type of statement.

(D) "If it rains, the ground is wet."

This is a conditional statement expressing a logical implication about a real-world scenario. It is not a mathematical statement involving an integer variable $n$. It cannot be proven using PMI.

Among the given options, only option (C) is a statement formulated as a property or formula depending on a positive integer $n$ which is suitable for proof by the Principle of Mathematical Induction.

The correct option is (C) For every positive integer $n$, $1 \cdot 2 + 2 \cdot 3 + ... + n(n+1) = \frac{n(n+1)(n+2)}{3}$.

The statement is $P(n): n^2 + 1 > 2n$. This is equivalent to $(n-1)^2 > 0$.

Let's check the truth value of $P(n)$ for small positive integers:

For $n=1$: $P(1)$ is $(1-1)^2 > 0$, which is $0 > 0$. This is false.

For $n=2$: $P(2)$ is $(2-1)^2 > 0$, which is $1 > 0$. This is true.

For $n=3$: $P(3)$ is $(3-1)^2 > 0$, which is $4 > 0$. This is true.

Now let's evaluate the given options:

(A) Base case: $P(1)$ is true because $1^2 + 1 > 2(1)$, i.e., $2 > 2$, which is false. This statement accurately shows that the inequality $2 > 2$ resulting from $P(1)$ is false.

(B) Base case: $P(1)$ is true because $1^2 + 1 > 2(1)$, i.e., $2 \geq 2$, which is true. This statement incorrectly claims $P(1)$ is true and uses the wrong inequality ($\geq$ instead of $>$).

(C) Base case: $P(2)$ is true because $2^2 + 1 > 2(2)$, i.e., $5 > 4$, which is true. This statement correctly verifies $P(2)$ is true.

(D) Base case: $P(n)$ is true for $n=1, 2, 3$. Since $P(1)$ is false, this statement is false.

We are looking for the statement that is NOT true for verifying $P(n)$ using PMI.

Option (A) correctly indicates that $P(1)$ is false.

Option (C) correctly indicates that $P(2)$ is true.

Options (B) and (D) make false claims. Option (B) specifically makes a false claim about the base case $n=1$ and uses an incorrect calculation ($2 \geq 2$ instead of $2 > 2$). Option (D) makes a false claim about the truth of $P(n)$ for $n=1, 2, 3$.

Option (B) is the most direct incorrect statement about the base case verification using the correct inequality from $P(n)$. It asserts $P(1)$ is true, which is false, and its justification is flawed.

The correct option is (B) Base case: $P(1)$ is true because $1^2 + 1 > 2(1)$, i.e., $2 \geq 2$, which is true.

The Principle of Mathematical Induction (PMI) is a proof technique used to establish that a given statement $P(n)$ is true for all integers $n$ in a specific range, typically starting from a particular integer.

The problem states that two conditions for PMI are met:

1. The base case: $P(n)$ is true for $n=1$.

2. The inductive step: The truth of $P(k)$ implies the truth of $P(k+1)$ for all integers $k \geq 1$.

These two conditions, when combined, establish that the statement $P(n)$ is true for the initial integer specified in the base case ($n=1$) and for every subsequent integer. Starting from $n=1$, the implication $P(1) \implies P(2)$ means $P(2)$ is true. Then, $P(2) \implies P(3)$ means $P(3)$ is true, and so on. This chain of implications covers all integers $n$ that are greater than or equal to 1.

The set of integers $n$ for which $P(n)$ is proven true under these conditions is the set of all integers greater than or equal to 1.

Let's consider the options:

(A) All real numbers $n \geq 1$: PMI is a method for proving statements about integers, not all real numbers.

(B) All integers $n \geq 1$: This set includes $1, 2, 3, ...$, which is exactly the set of numbers covered by the induction with a base case at $n=1$ and the inductive step for $k \geq 1$.

(C) All integers $n > 1$: This set includes $2, 3, 4, ...$. It excludes the base case $n=1$, for which the statement is explicitly stated to be true.

(D) Only for $n=1, 2, 3, ...$ up to some large number: PMI proves the statement for the entire infinite sequence of integers starting from the base case, not just up to some finite number.

Based on the Principle of Mathematical Induction, if $P(1)$ is true and $P(k) \implies P(k+1)$ for all $k \geq 1$, then $P(n)$ is true for all integers $n \geq 1$.

The correct option is (B) All integers $n \geq 1$.

The Principle of Mathematical Induction (PMI) is a proof technique used to prove statements that are asserted to be true for all integers greater than or equal to some starting integer (the base case).

The principle is based on the structure of the set of integers. If we can show a statement $P(n)$ is true for an initial integer (e.g., $n=1$) and that if it is true for any integer $k$ (greater than or equal to the base case), it must also be true for the next integer $k+1$, then it is true for all integers starting from the base case.

Let's consider the options:

(A) Real numbers: Real numbers form a continuum and are not indexed by integers in the way required for standard induction.

(B) Complex numbers: Similar to real numbers, complex numbers in general are not the typical domain for PMI, though statements about sequences or series of complex numbers indexed by integers might use induction.

(C) Positive integers: The set of positive integers $\{1, 2, 3, ...\}$ is the most common domain for which statements are proven using PMI. The base case is often $n=1$.

(D) All integers: While variations or extensions of induction can sometimes be used to prove statements about all integers (positive, negative, and zero), the basic form of PMI (proving $P(n)$ for $n \geq n_0$) is most directly applicable to positive integers or subsets of integers bounded below.

In its most standard and fundamental application, the Principle of Mathematical Induction is used to prove statements about positive integers.

The correct option is (C) Positive integers.

The Principle of Mathematical Induction is used to prove that a statement $P(n)$ is true for all integers $n$ greater than or equal to some specific starting integer.

When proving $P(n)$ for all integers $n \geq m$, the induction starts from the smallest integer in this range, which is $m$.

The two steps of PMI for proving $P(n)$ for all $n \geq m$ are:

1. Base Case: Prove that the statement $P(n)$ is true for the smallest value in the set, which is $n=m$.

2. Inductive Step: Assume that the statement $P(k)$ is true for some arbitrary integer $k \geq m$ (the inductive hypothesis). Then, prove that the statement $P(k+1)$ is true.

Let's look at the options:

(A) 0: This would be the base case if proving for $n \geq 0$ (non-negative integers), but the range is $n \geq m$.

(B) 1: This would be the base case if proving for $n \geq 1$ (positive integers), but the range is $n \geq m$.

(C) $m$: This is the smallest integer in the specified range $n \geq m$. According to the principle, this is where the base case must be proven.

(D) $m+1$: This is the integer that follows $m$. This value is relevant in the inductive step (as $k+1$ when $k=m$), but the base case is for the starting value $m$.

Therefore, the base case should be proven for $n=m$ when proving $P(n)$ for all integers $n \geq m$.

The correct option is (C) $m$.

Let's analyze the Assertion (A) and the Reason (R).

Assertion (A): If the base case $P(1)$ is true and the inductive step (assuming $P(k)$ is true implies $P(k+1)$ is true) is proven, then $P(n)$ is true for all positive integers $n$.

This is a correct statement of the Principle of Mathematical Induction for the domain of positive integers $\{1, 2, 3, ...\}$, where the base case starts at $n=1$. If the statement holds for the first value ($n=1$) and the truth for any integer $k \geq 1$ guarantees the truth for the next integer $k+1$, then the statement must hold for $1, 2, 3, ...$ sequentially.

Therefore, Assertion (A) is true.

Reason (R): The Principle of Mathematical Induction is a valid proof technique for statements about natural numbers.

Natural numbers are typically understood as positive integers $\{1, 2, 3, ...\}$ or non-negative integers $\{0, 1, 2, 3, ...\}$. The Principle of Mathematical Induction is indeed a well-established and valid proof technique specifically designed for proving statements about such integer domains.

Therefore, Reason (R) is true.

Now let's consider if Reason (R) is the correct explanation of Assertion (A).

Assertion (A) describes the application of PMI (with base case $n=1$) to prove a statement for all positive integers. Reason (R) states that PMI itself is a valid technique for statements about natural numbers (which include positive integers).

The reason why the steps described in Assertion (A) lead to the conclusion that $P(n)$ is true for all positive integers is precisely because the Principle of Mathematical Induction is a valid method for such proofs on the set of natural numbers. Reason (R) provides the underlying justification for the process described in Assertion (A).

Therefore, Reason (R) is the correct explanation of Assertion (A).

Both A and R are true, and R is the correct explanation of A.

The correct option is (A) Both A and R are true and R is the correct explanation of A.

The statement is $P(n): n! > 3^n$. We check the options provided for the smallest positive integer $n$ for which $P(n)$ is true.

• For $n=1$: $1! = 1$, $3^1 = 3$. Is $1 > 3$? No. ($P(1)$ is false)
• For $n=2$: $2! = 2$, $3^2 = 9$. Is $2 > 9$? No. ($P(2)$ is false)
• For $n=3$: $3! = 6$, $3^3 = 27$. Is $6 > 27$? No. ($P(3)$ is false)
• For $n=4$: $4! = 24$, $3^4 = 81$. Is $24 > 81$? No. ($P(4)$ is false)

None of the options (1, 2, 3, 4) make the statement $n! > 3^n$ true. The smallest positive integer for which $n! > 3^n$ is true is $n=7$ ($7! = 5040$, $3^7 = 2187$, and $5040 > 2187$).

Given the multiple-choice format, it is highly probable that there is a typo in the question and the intended inequality was $n! > 2^n$. Let's check the options for this inequality:

• For $n=1$: $1! = 1$, $2^1 = 2$. Is $1 > 2$? No.
• For $n=2$: $2! = 2$, $2^2 = 4$. Is $2 > 4$? No.
• For $n=3$: $3! = 6$, $2^3 = 8$. Is $6 > 8$? No.
• For $n=4$: $4! = 24$, $2^4 = 16$. Is $24 > 16$? Yes. ($P(4)$ is true for $n! > 2^n$)

Assuming the question intended to ask for the smallest positive integer $n$ for which $n! > 2^n$ is true, the answer is $n=4$, which matches option (D).

Based on the provided options, and assuming a likely typo in the inequality, the correct option is (D) 4 (for the statement $n! > 2^n$).

The Standard Principle of Mathematical Induction (PMI) for positive integers starting from 1 is used to prove a statement $P(n)$ is true for all $n \in \{1, 2, 3, ...\}$. This involves proving the base case $P(1)$ and the inductive step ($P(k) \implies P(k+1)$ for $k \geq 1$).

Let's examine each statement to see if it can be proven for all $n \geq 1$ using standard PMI starting from $n=1$.

(A) The sum of the first $n$ odd numbers is $n^2$. ($P(n): 1 + 3 + ... + (2n-1) = n^2$)

Base case $n=1$: $1 = 1^2$. True. This statement can be proven by standard PMI starting from $n=1$.

(B) $n(n+1)(n+2)$ is divisible by 6 for all $n \geq 1$. ($P(n): n(n+1)(n+2)$ is divisible by 6)

Base case $n=1$: $1(2)(3) = 6$. 6 is divisible by 6. True. This statement can be proven by standard PMI starting from $n=1$.

(C) $1^3 + 2^3 + ... + n^3 = (\frac{n(n+1)}{2})^2$. ($P(n): \sum_{i=1}^{n} i^3 = (\frac{n(n+1)}{2})^2$)

Base case $n=1$: $1^3 = 1$. $(\frac{1(2)}{2})^2 = 1^2 = 1$. True. This statement can be proven by standard PMI starting from $n=1$.

(D) $n! > 2^n$ for all $n \geq 1$. ($P(n): n! > 2^n$)

Base case $n=1$: $1! = 1$, $2^1 = 2$. Is $1 > 2$? This is false. Since the base case $P(1)$ is false, this statement cannot be proven to be true for all positive integers $n \geq 1$ using standard PMI starting from $n=1$. (Note: This statement is actually true for $n \geq 4$, and can be proven by PMI starting with the base case $n=4$).

The statement that cannot be proven true for all positive integers $n \geq 1$ using standard PMI starting from the base case $n=1$ is option (D), because the base case $P(1)$ is false.

The correct option is (D) $n! > 2^n$ for all $n \geq 1$.

In the context of a proof by Mathematical Induction, the statement $P(n)$ is the specific property or assertion about the integer $n$ that is being proven.

The problem states that the advisor wants to prove using induction that "$A(n) > 1000 + 100n$ for all positive integers $n \geq 2$".

The formula for $A(n)$ is given as $A(n) = 1000 \times (1.1)^n$.

The statement to be proven is the inequality "$A(n) > 1000 + 100n$".

Substituting the formula for $A(n)$, the inequality is $1000 \times (1.1)^n > 1000 + 100n$.

Let's look at the options:

(A) $A(n) = 1000 \times (1.1)^n$: This is the definition of $A(n)$, not the statement being proven by induction.

(B) $A(n) > 1000 + 100n$: This is the inequality the advisor wants to prove about $A(n)$. This is the statement $P(n)$.

(C) $1000 \times (1.1)^n > 1000 + 100n$: This is the inequality the advisor wants to prove, with $A(n)$ replaced by its formula. This is equivalent to the statement $A(n) > 1000 + 100n$. This is also the statement $P(n)$.

(D) $(1.1)^n > 1 + 0.1n$: This inequality is obtained by dividing the inequality in option (C) by 1000. If $1000 \times (1.1)^n > 1000 + 100n$, then dividing by 1000 (a positive number) gives $(1.1)^n > \frac{1000}{1000} + \frac{100n}{1000}$, which is $(1.1)^n > 1 + 0.1n$. This is an equivalent statement, often easier to work with in the inductive step (related to Bernoulli's Inequality). It is also a valid formulation of $P(n)$.

Options (B), (C), and (D) all represent the statement $P(n)$ that is being proven. However, option (B) uses the notation $A(n)$ directly as given in the problem description of the statement to be proven, while (C) replaces $A(n)$ with its explicit formula, and (D) is an equivalent simplified inequality. In the context of defining the statement $P(n)$, the most direct representation of the statement the advisor wants to prove is (B) or (C).

The question asks for "the statement $P(n)$". Options (B) and (C) are identical statements. Option (D) is an equivalent statement. Typically, $P(n)$ is defined as the direct assertion being made about $n$. In this case, the advisor wants to prove $A(n) > 1000 + 100n$. Option (B) states this directly using the notation $A(n)$. Option (C) states the same thing using the explicit formula for $A(n)$. Option (D) is an algebraically manipulated version.

Option (B) is the statement about $A(n)$ as mentioned in the problem description ("prove ... that $A(n) > 1000 + 100n$"). Option (C) is that statement with the formula for $A(n)$ substituted. Option (D) is an algebraic simplification of the statement.

Given the options, (B) or (C) are the most direct interpretations of $P(n)$. Since (C) explicitly shows the formula, it is also a strong candidate. Option (B) uses the defined term $A(n)$. Without further context on how $P(n)$ should be expressed (using defined terms or explicit formulas), both (B) and (C) are mathematically equivalent and represent the statement being proven.

However, the options are distinct choices. Let's consider the phrasing "What is the statement $P(n)$ in this case?". It is the assertion made about $n$. That assertion is "$A(n) > 1000 + 100n$". Option (B) is exactly this assertion. Option (C) is the same assertion written using the formula for $A(n)$. Option (D) is an equivalent assertion after algebraic manipulation.

In many texts, $P(n)$ is written using the most explicit form of the expression. Therefore, (C) is a very common way to write $P(n)$. Option (B) uses the defined variable $A(n)$, which is also valid. Option (D) is also valid as an equivalent statement.

Let's re-read the question carefully: "The advisor wants to prove using induction that $A(n) > 1000 + 100n$ for all positive integers $n \geq 2$." The statement being proven is "$A(n) > 1000 + 100n$". This matches option (B).

While (C) and (D) are equivalent forms, option (B) is the direct representation of the inequality stated in the problem using the variable $A(n)$.

The correct option is (B) $A(n) > 1000 + 100n$.

The problem states that the advisor wants to prove the statement $P(n): A(n) > 1000 + 100n$ for all positive integers $n \geq 2$.

When proving a statement $P(n)$ for all integers $n \geq m$, the base case is to verify the statement for the smallest value in the range, which is $n=m$. In this case, the range is $n \geq 2$, so the smallest value is $n=2$.

The base case is $P(2)$. We need to verify the statement $P(n)$ for $n=2$. Substituting $n=2$ into $P(n): A(n) > 1000 + 100n$, we get:

$P(2): A(2) > 1000 + 100(2)$

$P(2): A(2) > 1000 + 200$

$P(2): A(2) > 1200$

Now let's calculate $A(2)$ using the given formula $A(n) = 1000 \times (1.1)^n$:

$A(2) = 1000 \times (1.1)^2$

$A(2) = 1000 \times (1.1 \times 1.1)$

$A(2) = 1000 \times 1.21$

$A(2) = 1210$

So, the statement $P(2)$ is $1210 > 1200$, which is true.

Let's look at the options provided for the base case that needs to be verified:

(A) $P(1): A(1) > 1000 + 100(1)$: This is checking the base case for $n=1$. The problem specifies $n \geq 2$, so $n=1$ is not the starting point.

(B) $P(2): A(2) > 1000 + 100(2)$: This correctly identifies $n=2$ as the base case and writes the statement $P(2)$ using the notation $A(n)$.

(C) $P(1): 1000 \times 1.1 > 1100$: This is checking the base case for $n=1$, using the formula for $A(1)$. $A(1) = 1000 \times (1.1)^1 = 1100$. The right side is $1000 + 100(1) = 1100$. So $P(1)$ is $1100 > 1100$, which is false. More importantly, the base case is not $n=1$.

(D) $P(2): 1000 \times (1.1)^2 > 1200$: This correctly identifies $n=2$ as the base case and writes the statement $P(2)$ using the formula for $A(2)$ and the value of the right side $1000 + 100(2) = 1200$. This is the statement $1210 > 1200$.

Both options (B) and (D) correctly identify the base case as $n=2$. Option (B) expresses the statement $P(2)$ using $A(2)$ notation, while option (D) expresses it using the explicit formula for $A(2)$ and the calculated value of the right side. Both represent the base case $P(2)$. However, option (D) shows the numerical values involved in the inequality that needs to be verified, which is a step towards the verification process itself. Option (B) simply states the inequality $P(2)$ in terms of $A(2)$.

Given the options, (D) provides the specific numerical inequality derived from $P(2)$ that must be checked for truth, making it a more complete description of the verification step than just stating $P(2)$ as in (B).

The base case to be verified is $P(2)$, which is $A(2) > 1000 + 100(2)$, or $1000 \times (1.1)^2 > 1200$.

The correct option is (D) $P(2): 1000 \times (1.1)^2 > 1200$.

The statement is $P(n): 1000 \times (1.1)^n > 1000 + 100n$. We are proving this for $n \geq 2$.

In the inductive step, we assume $P(k)$ is true for some integer $k \geq 2$ (the base case starting value). The inductive hypothesis is:

We need to prove that $P(k+1)$ is true, which is the statement $P(n)$ with $n$ replaced by $k+1$:

$P(k+1): 1000 \times (1.1)^{k+1} > 1000 + 100(k+1)$

To prove $P(k+1)$, we typically start with the left side of the inequality $P(k+1)$ and use the inductive hypothesis (equation (i)) to show it is greater than the right side of $P(k+1)$.

The left side of $P(k+1)$ is $1000 \times (1.1)^{k+1}$.

The right side of $P(k+1)$ is $1000 + 100(k+1) = 1000 + 100k + 100$.

Let's look at the options for the expression we should consider:

(A) $1000 \times (1.1)^{k+1}$: This is the left side of the inequality we need to prove for $P(k+1)$. We start manipulating this expression in the inductive step.

(B) $1000 + 100(k+1)$: This is the right side of the inequality we need to prove for $P(k+1)$. We need to show that the expression in (A) is greater than this expression.

(C) $1000 \times (1.1)^k \times 1.1$: This is just a rewritten form of the expression in (A), using exponent rules $(1.1)^{k+1} = (1.1)^k \times (1.1)^1$. This form is particularly useful because it contains the term $1000 \times (1.1)^k$, which appears in our inductive hypothesis (equation (i)). We would typically start with $1000 \times (1.1)^{k+1}$, rewrite it as $1000 \times (1.1)^k \times 1.1$, and then use the hypothesis $1000 \times (1.1)^k > 1000 + 100k$.

Multiplying the inductive hypothesis by 1.1 (since 1.1 > 0):

$1.1 \times [1000 \times (1.1)^k] > 1.1 \times [1000 + 100k]$

$1000 \times (1.1)^{k+1} > 1100 + 110k$

Now we need to show that $1100 + 110k > 1000 + 100k + 100$.

This simplifies to $1100 + 110k > 1100 + 100k$.

Subtracting $1100 + 100k$ from both sides, we need to show $10k > 0$. Since $k \geq 2$, this is true.

So, $1000 \times (1.1)^{k+1} > 1100 + 110k > 1100 + 100k = 1000 + 100(k+1)$.

This shows that $P(k+1)$ is true.

(D) All of the above are relevant to the inductive step: As shown in the process above, we need to consider the left side of $P(k+1)$ (options A and C are forms of this), and we need to compare it to the right side of $P(k+1)$ (option B). Therefore, all three expressions are relevant when performing the inductive step.

The correct option is (D) All of the above are relevant to the inductive step.

The statement is $P(n): 2^n > n^2$. We need to find which of the given statements is a true statement regarding verifying this using PMI.

Let's check the truth value of $P(n)$ for the first few positive integers:

• For $n=1$: $P(1): 2^1 > 1^2 \implies 2 > 1$. This is true.
• For $n=2$: $P(2): 2^2 > 2^2 \implies 4 > 4$. This is false.
• For $n=3$: $P(3): 2^3 > 3^2 \implies 8 > 9$. This is false.
• For $n=4$: $P(4): 2^4 > 4^2 \implies 16 > 16$. This is false.
• For $n=5$: $P(5): 2^5 > 5^2 \implies 32 > 25$. This is true.
• For $n=6$: $P(6): 2^6 > 6^2 \implies 64 > 36$. This is true.

The statement $P(n)$ is true for $n=1$, false for $n=2, 3, 4$, and true for $n \geq 5$.

To prove the statement $P(n)$ for all integers $n$ where it is true, we need to find the smallest integer $m$ such that $P(m)$ is true, and $P(k) \implies P(k+1)$ for all $k \geq m$. In this case, the statement is not true for all positive integers starting from 1 because it fails for $n=2, 3, 4$. The statement is true for $n=1$ and for all integers $n \geq 5$. If we want to prove it for all integers where it is true, we would likely need to handle $n=1$ separately or prove it for $n \geq 5$ using a base case at $n=5$.

Let's evaluate the given options:

(A) $P(1)$ is true ($2^1 > 1^2 \implies 2 > 1$). This statement claims $P(1)$ is true and provides the correct calculation ($2 > 1$) and its truth value (True). This is a true statement about $P(1)$.

(B) $P(2)$ is true ($2^2 > 2^2 \implies 4 > 4$), which is false. This statement claims $P(2)$ is true, but the calculation ($4 > 4$) is false. So the claim "P(2) is true" is false. The entire statement is asserting something false about $P(2)$. This is a false statement.

(C) $P(3)$ is true ($2^3 > 3^2 \implies 8 > 9$), which is false. This statement claims $P(3)$ is true, but the calculation ($8 > 9$) is false. So the claim "P(3) is true" is false. The entire statement is asserting something false about $P(3)$. This is a false statement.

(D) The base case needs to start at $n=5$ because $P(4)$ is true ($2^4 > 4^2 \implies 16 > 16$) is false, but $P(5)$ is true ($2^5 > 5^2 \implies 32 > 25$). This statement claims that the base case needs to start at $n=5$. It correctly states that $P(4)$ is false ($16>16$ is false) and $P(5)$ is true ($32>25$). Since $P(n)$ is false for $n=2, 3, 4$ and true for $n=5$, the smallest integer $m$ such that $P(n)$ is true for all $n \geq m$ (and where the inductive step $P(k) \implies P(k+1)$ might hold for $k \geq m$) is $m=5$. Therefore, to prove $P(n)$ for $n \geq 5$, the base case should start at $n=5$. This is a true statement about the appropriate base case for proving $P(n)$ for $n \geq 5$ using PMI, given the truth values of $P(n)$.

We are looking for a true statement among the options. Options (A) and (D) are true statements. Options (B) and (C) are false statements.

If the question implies finding a starting point for induction to prove the statement where it is true for all subsequent integers, option (D) is a relevant and true statement about the verification process for $P(n)$. Option (A) is simply a true statement about the truth of $P(1)$, but $P(1)$ being true alone is not sufficient for standard PMI for all $n \geq 1$ if the inductive step fails for some $k \geq 1$ (which it does between $k=1$ and $k=2$).

The phrasing "Which of the following is a true statement for verifying this using PMI?" suggests looking at correct facts related to applying PMI to $P(n)$. Both (A) and (D) provide correct facts.

However, typically, a question like this in the context of PMI exercises is guiding towards understanding where the induction proof should start for the main range where the statement holds. Since $P(n)$ fails for $n=2,3,4$, induction starting at $n=1$ for all $n \geq 1$ will not work. The statement holds for $n=1$ and for $n \geq 5$. To prove it for the range $n \geq 5$, the base case must be $n=5$. Option (D) correctly identifies this.

Given the structure of the problem and options, option (D) provides a more complete insight into applying PMI to this specific statement across its domain of truth ($n \geq 5$) compared to option (A) which only states the truth of the statement at $n=1$ without addressing the subsequent falsity.

The correct option is (D) The base case needs to start at $n=5$ because $P(4)$ is true ($2^4 > 4^2 \implies 16 > 16$) is false, but $P(5)$ is true ($2^5 > 5^2 \implies 32 > 25$).

The question asks about the crucial property of natural numbers used when proving the formula for the sum of the first $n$ natural numbers using the Principle of Mathematical Induction (PMI).

The statement to be proven is $P(n): 1 + 2 + 3 + ... + n = \frac{n(n+1)}{2}$. This is proven for all positive integers $n$ (assuming natural numbers start from 1).

The steps of the proof using PMI are:

1. Base Case: Prove $P(1)$ is true. $1 = \frac{1(1+1)}{2}$. This step uses basic arithmetic properties of natural numbers.

2. Inductive Step: Assume $P(k)$ is true for some positive integer $k$, i.e., $1 + 2 + ... + k = \frac{k(k+1)}{2}$. Prove $P(k+1)$ is true, i.e., $1 + 2 + ... + k + (k+1) = \frac{(k+1)((k+1)+1)}{2}$. This step involves manipulating the equation, using arithmetic, and relies on the fundamental structure of natural numbers which allows for this step-by-step implication.

The structure of the proof itself, where proving for $k$ implies proving for $k+1$ and starting from a base case covers the entire set, is based on the Principle of Mathematical Induction.

Let's consider the options:

(A) Commutative property of addition ($a+b = b+a$): While used in arithmetic, it is not the core property of natural numbers that makes PMI work.

(B) Associative property of addition ($(a+b)+c = a+(b+c)$): Also used in arithmetic, not the core principle behind induction.

(C) Principle of Mathematical Induction: This is itself a principle about the structure of natural numbers (or integers greater than or equal to a base integer). It states that if a property holds for the first element and if holding for any element implies holding for the next, then the property holds for all elements in the sequence. This is the fundamental principle that allows the proof to extend from the base case to all subsequent natural numbers.

(D) Distributive property ($a(b+c) = ab+ac$): Used in algebraic manipulation, but not the crucial property of natural numbers that underpins the induction process itself.

The proof relies on arithmetic properties (like commutative, associative, distributive) during the calculations within the base and inductive steps. However, the ability to conclude that the statement is true for *all* natural numbers after proving the base case and the inductive step comes directly from the Principle of Mathematical Induction itself, which is a property of the structure of the set of natural numbers (or integers). Therefore, PMI is the crucial property that allows this type of proof.

The correct option is (C) Principle of Mathematical Induction.

The statement is $P(n): n(n+1)(n+5)$ is a multiple of 3 for all positive integers $n$.

Let's analyze each part of the induction proof and match it to the descriptions provided.

(i) Base Case $P(1)$:

The base case involves checking the statement $P(n)$ for the smallest value of $n$ in the domain, which is $n=1$ for positive integers. So, we need to check if $1(1+1)(1+5)$ is a multiple of 3.

$1(1+1)(1+5) = 1 \cdot 2 \cdot 6 = 12$.

Is 12 a multiple of 3? Yes, $12 = 3 \times 4$.

This matches description (c) $1(1+1)(1+5) = 1 \cdot 2 \cdot 6 = 12$, which is a multiple of 3.

So, (i) - (c).

(ii) Inductive Hypothesis:

The inductive hypothesis is the assumption that the statement $P(k)$ is true for some arbitrary positive integer $k$.

$P(k)$ is the statement "$k(k+1)(k+5)$ is a multiple of 3".

This matches description (a) $k(k+1)(k+5)$ is a multiple of 3.

So, (ii) - (a).

(iii) Expression for $P(k+1)$:

The statement $P(k+1)$ is obtained by replacing $n$ with $k+1$ in the statement $P(n)$.

$P(k+1)$ is the statement "$(k+1)((k+1)+1)((k+1)+5)$ is a multiple of 3".

The expression for $P(k+1)$ is $(k+1)((k+1)+1)((k+1)+5)$, which simplifies to $(k+1)(k+2)(k+6)$.

This matches description (d) $(k+1)((k+1)+1)((k+1)+5)$.

So, (iii) - (d).

(iv) Goal of Inductive Step:

The goal of the inductive step is to prove that $P(k+1)$ is true, assuming $P(k)$ is true. So, we need to show that the expression for $P(k+1)$ (which is $(k+1)(k+2)(k+6)$) is a multiple of 3.

This matches description (b) Show $(k+1)(k+2)(k+6)$ is a multiple of 3.

So, (iv) - (b).

Matching the parts:

• (i) - (c)
• (ii) - (a)
• (iii) - (d)
• (iv) - (b)

Let's check the options provided based on this matching:

(A) (i)-(a), (ii)-(b), (iii)-(c), (iv)-(d) - Incorrect

(B) (i)-(c), (ii)-(a), (iii)-(d), (iv)-(b) - Correct

(C) (i)-(c), (ii)-(d), (iii)-(a), (iv)-(b) - Incorrect

(D) (i)-(a), (ii)-(c), (iii)-(d), (iv)-(b) - Incorrect

The correct matching is (i)-(c), (ii)-(a), (iii)-(d), (iv)-(b).

The correct option is (B) (i)-(c), (ii)-(a), (iii)-(d), (iv)-(b).

The statement is $P(n): 2^{3n} - 1$ is divisible by 7 for every positive integer $n$.

In the inductive step, we assume $P(k)$ is true for some positive integer $k$.

Inductive Hypothesis: $P(k)$ is true, meaning $2^{3k} - 1$ is divisible by 7. This can be written as:

From this, we can write $2^{3k} = 7m + 1$.

We need to prove $P(k+1)$ is true, meaning $2^{3(k+1)} - 1$ is divisible by 7.

We consider the expression $2^{3(k+1)} - 1$ and try to show it is divisible by 7, using the inductive hypothesis.

Let's manipulate the expression $2^{3(k+1)} - 1$ step by step:

First, simplify the exponent:

$2^{3(k+1)} - 1 = 2^{3k + 3} - 1$

Using exponent rules ($a^{x+y} = a^x \cdot a^y$):

$= 2^{3k} \cdot 2^3 - 1$

This matches option (A).

Calculate the value of $2^3$: $2^3 = 2 \times 2 \times 2 = 8$.

Substitute this value into the expression:

$= 8 \cdot 2^{3k} - 1$

This matches option (B).

Now, we use the inductive hypothesis $2^{3k} = 7m + 1$. Substitute this into the expression from option (B):

$= 8(7m + 1) - 1$

Distribute the 8:

$= 56m + 8 - 1$

Combine the constants:

$= 56m + 7$

Factor out 7:

$= 7(8m + 1)$

Since $m$ is an integer, $8m+1$ is also an integer. Therefore, $2^{3(k+1)} - 1$ is a multiple of 7, which means it is divisible by 7. This matches the calculation shown in option (C).

Let's look at the options provided for what the expression $2^{3(k+1)} - 1$ is equal to:

(A) $2^{3k} \cdot 2^3 - 1$: This is a correct step in simplifying the expression.

(B) $8 \cdot 2^{3k} - 1$: This is also a correct step, resulting from calculating $2^3=8$. It is equivalent to (A).

(C) $8(7m+1) - 1 = 56m + 8 - 1 = 56m + 7 = 7(8m+1)$: This shows the process of substituting the inductive hypothesis and manipulating the expression to show it is a multiple of 7. The intermediate steps and the final result $7(8m+1)$ are correct and show that the expression is divisible by 7.

(D) All of the above steps lead to the conclusion: Options (A), (B), and (C) represent sequential steps in the manipulation of the expression $2^{3(k+1)} - 1$ to use the inductive hypothesis and show divisibility by 7. Starting from $2^{3(k+1)} - 1$, we can rewrite it as in (A), then as in (B), and finally use the inductive hypothesis as shown in the calculation in (C) to reach the form $7 \times (\text{an integer})$, which demonstrates divisibility by 7. Therefore, all three expressions/steps are part of the process that leads to the conclusion that $P(k+1)$ is true.

The question asks what the expression $2^{3(k+1)} - 1$ is equal to. Option (A), (B), and the first part of the calculation in (C) ($8(7m+1)-1$) are equal to the original expression $2^{3(k+1)} - 1$ through algebraic manipulation and substitution of the inductive hypothesis. Option (C) goes further to show the result is a multiple of 7.

Option (D) states that all the above steps lead to the conclusion. This is true. Options (A), (B), and the process shown in (C) are valid algebraic steps in the inductive proof.

Given the choices, option (D) is the most comprehensive answer, indicating that all the listed expressions and calculations are relevant and form part of the inductive step process that leads to the proof of $P(k+1)$.

The correct option is (D) All of the above steps lead to the conclusion.

The statement given is $P(n): (1+x)^n \geq 1+nx$ for all positive integers $n$ (as $\mathbb{N}$ often denotes $\{1, 2, 3, ...\}$) and for a real number $x > -1$.

Let's recall some common mathematical inequalities and theorems:

(A) Binomial Theorem: This theorem provides a formula for expanding powers of a binomial $(a+b)^n$: $(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$. It's an equality, not an inequality of the form given.

(B) Cauchy-Schwarz Inequality: This inequality relates the sum of the products of real numbers to the products of their sums of squares: $(\sum_{i=1}^n a_i b_i)^2 \leq (\sum_{i=1}^n a_i^2)(\sum_{i=1}^n b_i^2)$. This is not the inequality given.

(C) Bernoulli's Inequality: This inequality states that for any integer $n \geq 0$ and any real number $x > -1$, $(1+x)^n \geq 1+nx$. If $n=0$, $(1+x)^0 = 1$ and $1+0x = 1$, so $1 \geq 1$ (true). If $n \geq 1$ and $x > -1$, the inequality holds. The statement in the question matches Bernoulli's Inequality, particularly for positive integers $n \in \mathbb{N}$.

(D) Arithmetic Mean-Geometric Mean Inequality (AM-GM Inequality): This inequality states that for a set of non-negative real numbers, the arithmetic mean is greater than or equal to the geometric mean: $\frac{a_1 + a_2 + ... + a_n}{n} \geq \sqrt[n]{a_1 a_2 ... a_n}$. This is not the inequality given.

Comparing the given statement $(1+x)^n \geq 1+nx$ with the definitions, it precisely matches the statement of Bernoulli's Inequality for $n \in \mathbb{N}$ and $x > -1$. The condition $x > -1$ is important because if $x \leq -1$ and $n$ is odd, the inequality may not hold (e.g., $n=3, x=-2$: $(1-2)^3 = (-1)^3 = -1$, $1+3(-2) = 1-6 = -5$, so $-1 \geq -5$ is true. But if $n=2, x=-3$: $(1-3)^2 = (-2)^2 = 4$, $1+2(-3) = 1-6 = -5$, so $4 \geq -5$ is true. However, the standard proof by induction requires $x > -1$). The condition $x > -1$ ensures that $1+x$ is positive, which is often used in the inductive step (multiplying by $1+x$). If $x=0$ or $x=-1$, the inequality becomes an equality. If $x > -1$ and $x \neq 0$, the inequality is strict for $n > 1$.

The statement is known as Bernoulli's Inequality.

The correct option is (C) Bernoulli's Inequality.

The Principle of Mathematical Induction (PMI) for proving a statement $P(n)$ is true for all positive integers $n \geq 1$ consists of two essential steps:

1. Base Case: Prove that $P(1)$ is true.

2. Inductive Step: Assume $P(k)$ is true for some arbitrary positive integer $k \geq 1$ (this is the inductive hypothesis), and then prove that $P(k+1)$ is true using this assumption.

If both of these steps are successfully completed, the Principle of Mathematical Induction guarantees that $P(n)$ is true for all positive integers $n \geq 1$.

Let's evaluate the given options:

(A) Proving $P(1)$ is true: This is the necessary and crucial first step (Base Case) in PMI for proving the statement for all positive integers starting from 1.

(B) Assuming $P(k)$ is true for some positive integer $k \geq 1$: This is the necessary assumption (Inductive Hypothesis) made at the beginning of the Inductive Step.

(C) Proving $P(k+1)$ is true based on the assumption that $P(k)$ is true: This is the necessary second part of the Inductive Step. Showing that the inductive hypothesis implies the truth of the statement for the next integer is fundamental to the proof.

(D) Checking if $P(n)$ is true for a few values like $n=2, 3, 4$ after the base case: Checking specific values of $n$ can be helpful for understanding the statement, forming a conjecture, or gaining confidence, but it is NOT a formal, valid step in the *proof* by PMI. The inductive step ensures that if $P(k)$ is true, then $P(k+1)$ is true for all $k \geq 1$. This logical implication covers all subsequent integers from the base case without needing to check them individually. Checking a few values after the base case does not add to the logical validity of the induction proof; the validity comes solely from the successful completion of the base case and the general inductive step.

Therefore, checking a few values of $n$ after the base case is not a required or valid step within the formal structure of a PMI proof.

The correct option is (D) Checking if $P(n)$ is true for a few values like $n=2, 3, 4$ after the base case.

The statement is $P(n): 1 \cdot 1! + 2 \cdot 2! + ... + n \cdot n! = (n+1)! - 1$.

The left-hand side (LHS) of $P(n)$ is the sum $1 \cdot 1! + 2 \cdot 2! + ... + n \cdot n!$. This is a sum of $n$ terms, where the $i$-th term is $i \cdot i!$.

The statement $P(k)$ is obtained by replacing $n$ with $k$: $1 \cdot 1! + 2 \cdot 2! + ... + k \cdot k! = (k+1)! - 1$. The LHS of $P(k)$ is the sum of the first $k$ terms: $1 \cdot 1! + 2 \cdot 2! + ... + k \cdot k!$.

The statement $P(k+1)$ is obtained by replacing $n$ with $k+1$. The LHS of $P(k+1)$ is the sum of the first $k+1$ terms of the series: $1 \cdot 1! + 2 \cdot 2! + ... + (k+1) \cdot (k+1)!$.

This sum consists of the sum of the first $k$ terms plus the $(k+1)$-th term.

LHS of $P(k+1) = \underbrace{1 \cdot 1! + 2 \cdot 2! + ... + k \cdot k!}_{\text{LHS of } P(k)} + (k+1) \cdot (k+1)!$

Let's look at the options:

(A) $1 \cdot 1! + 2 \cdot 2! + ... + k \cdot k!$: This is the left-hand side of $P(k)$, not $P(k+1)$.

(B) $1 \cdot 1! + 2 \cdot 2! + ... + k \cdot k! + (k+1) \cdot (k+1)!$: This is the sum of the first $k+1$ terms, which is the left-hand side of $P(k+1)$.

(C) $(k+1) \cdot (k+1)!$: This is just the $(k+1)$-th term of the series, not the entire sum on the left side of $P(k+1)$.

(D) $(k+2)! - 1$: This is the right-hand side of the statement $P(k+1)$ (obtained by replacing $n$ with $k+1$ in $(n+1)! - 1$), not the left-hand side.

The left-hand side of the statement $P(k+1)$ is the sum of the series up to the $(k+1)$-th term.

The correct option is (B) $1 \cdot 1! + 2 \cdot 2! + ... + k \cdot k! + (k+1) \cdot (k+1)!$.

The Principle of Mathematical Induction is used to prove that a statement $P(n)$ is true for all integers $n$ greater than or equal to some specific starting integer $m$.

When proving $P(n)$ for all integers $n \geq 5$, the domain of the statement is $\{5, 6, 7, 8, ...\}$.

The base case of a PMI proof is always the verification of the statement $P(n)$ for the smallest integer in the specified domain.

In this case, the smallest integer in the domain $n \geq 5$ is $n=5$.

Therefore, the base case should be $P(5)$.

Let's look at the options:

(A) 1: This would be the base case if proving for $n \geq 1$.

(B) 4: This is less than the starting value 5.

(C) 5: This is the smallest integer in the range $n \geq 5$, and is the correct base case.

(D) 6: This is greater than the starting value 5; it would be part of the inductive step, not the base case.

The correct option is (C) 5.

The statement is $P(n)$: "A set with $n$ elements has $2^n$ subsets". We are considering the base case $P(1)$.

The base case $P(1)$ means evaluating the statement $P(n)$ for $n=1$.

Substitute $n=1$ into the statement $P(n)$:

"A set with $1$ element has $2^1$ subsets".

Let's verify this. Consider a set with 1 element, for example, $\{a\}$. The subsets of this set are $\emptyset$ and $\{a\}$. There are exactly 2 subsets. And $2^1 = 2$. So, the statement $P(1)$ is true.

Let's look at the options:

(A) A set with 1 element has $2^1=2$ subsets. This is the correct statement for $P(1)$ and confirms its truth.

(B) The empty set has $2^0=1$ subset. This describes the statement $P(0)$. If the domain included $n=0$, this would be the base case $P(0)$. However, the question asks about $P(1)$.

(C) A set with 2 elements has $2^2=4$ subsets. This describes the statement $P(2)$.

(D) A set with $n$ elements has $2^n$ subsets for $n=1$. This describes what $P(1)$ asserts by referring back to the general form $P(n)$ for the case $n=1$. While technically true, option (A) directly states the meaning of $P(1)$ by substituting the value of $n$ and calculating $2^1$. Option (A) is a more direct and informative description of what $P(1)$ means in this context.

Option (A) explicitly evaluates the claim for $n=1$. Option (D) is a formal way of saying the same thing but doesn't show the calculation of $2^1$. In the context of a base case verification, evaluating the expression is usually part of the statement, as shown in option (A).

The correct option is (A) A set with 1 element has $2^1=2$ subsets.

We need to identify which of the given divisibility statements can be proven true for all positive integers $n$ (assuming $\mathbb{N} = \{1, 2, 3, ...\}$) using the Principle of Mathematical Induction (PMI).

For a statement $P(n)$ to be provable by PMI for all $n \geq 1$, it must first be true for the base case $n=1$, and the inductive step ($P(k) \implies P(k+1)$ for $k \geq 1$) must hold.

Let's check each statement for $n=1$ and consider whether the inductive step seems likely to hold.

(A) $P(n): n^2 + 1$ is divisible by 3.

Base case $n=1$: $P(1): 1^2 + 1 = 2$. Is 2 divisible by 3? No. $P(1)$ is false. Since the base case is false, this statement cannot be true for all $n \in \mathbb{N}$, and thus cannot be proven for all $n \in \mathbb{N}$ by PMI starting at $n=1$.

(B) $P(n): n^3 + 2n$ is divisible by 3.

Base case $n=1$: $P(1): 1^3 + 2(1) = 1 + 2 = 3$. Is 3 divisible by 3? Yes. $P(1)$ is true.

Let's consider the inductive step. Assume $P(k)$ is true for some $k \geq 1$, i.e., $k^3 + 2k$ is divisible by 3. We need to show $P(k+1)$ is true, i.e., $(k+1)^3 + 2(k+1)$ is divisible by 3.

$(k+1)^3 + 2(k+1) = (k^3 + 3k^2 + 3k + 1) + (2k + 2)$

$= k^3 + 3k^2 + 5k + 3$

$= (k^3 + 2k) + 3k^2 + 3k + 3$

$= (k^3 + 2k) + 3(k^2 + k + 1)$

By the inductive hypothesis, $k^3 + 2k$ is divisible by 3. The term $3(k^2 + k + 1)$ is clearly divisible by 3. The sum of two numbers divisible by 3 is also divisible by 3. Thus, $(k+1)^3 + 2(k+1)$ is divisible by 3. The inductive step holds.

Since $P(1)$ is true and $P(k) \implies P(k+1)$ for $k \geq 1$, this statement can be proven by PMI for all $n \in \mathbb{N}$.

(C) $P(n): 2^n + 1$ is divisible by 3.

Base case $n=1$: $P(1): 2^1 + 1 = 3$. Is 3 divisible by 3? Yes. $P(1)$ is true.

Let's check for $n=2$: $P(2): 2^2 + 1 = 4 + 1 = 5$. Is 5 divisible by 3? No. $P(2)$ is false.

Since $P(n)$ is not true for all $n \in \mathbb{N}$ (it fails for $n=2$), this statement cannot be proven by PMI for all $n \in \mathbb{N}$ starting at $n=1$. (Note: $2^n+1$ is divisible by 3 if and only if $n$ is odd). The inductive step would fail for even $k$.

(D) $P(n): n!$ is divisible by $n$.

Base case $n=1$: $P(1): 1! = 1$. Is 1 divisible by 1? Yes. $P(1)$ is true.

Let's consider the statement for $n > 1$. $n! = n \times (n-1) \times ... \times 1$. For $n > 1$, $n!$ has a factor of $n$. By definition of divisibility, if $n!$ has a factor of $n$, then $n!$ is divisible by $n$. This is true for all integers $n > 1$. So, the statement is true for all $n \geq 1$.

Can it be proven by PMI? Yes. We've shown $P(1)$ is true. Assume $P(k)$ is true for $k \geq 1$, i.e., $k!$ is divisible by $k$. We need to show $(k+1)!$ is divisible by $k+1$.

$(k+1)! = (k+1) \times k!$. Since $(k+1)!$ has a factor of $(k+1)$, it is divisible by $k+1$. The inductive step holds for $k \geq 1$.

Since $P(1)$ is true and $P(k) \implies P(k+1)$ for $k \geq 1$, this statement can be proven by PMI for all $n \in \mathbb{N}$.

We have found that statements (B) and (D) can be proven by PMI for all $n \in \mathbb{N}$. There seems to be an issue with the question or options, as multiple options are provable by PMI for all $n \in \mathbb{N}$.

Let's re-examine the question and options. All options are divisibility statements. We are asked which one *can* be proven. It implies only one of them can be proven for all $n \in \mathbb{N}$ using standard PMI (starting from $n=1$).

Statement (A) is false for $n=1$. Statement (C) is false for $n=2$. So (A) and (C) are not provable for all $n \in \mathbb{N}$.

Statements (B) and (D) are both true for all $n \in \mathbb{N}$ and provable by PMI. This suggests there might be an error in the question or options, or perhaps a nuance is intended.

Let's double-check the standard examples of PMI. Summation formulas and divisibility rules like the one in (B) are very common examples. The statement in (D) ($n!$ is divisible by $n$) is true for all $n \geq 1$. For $n=1$, $1! = 1$ is divisible by 1. For $n>1$, $n! = n \times (n-1)!$, which is clearly divisible by $n$. The inductive proof we sketched is valid.

It is possible that the intended question is asking which statement is a *typical* or *non-trivial* application of PMI among the options, or that there is a subtle reason why one is preferred over the other in the context from which this question was taken. However, based purely on the mathematical validity of the proof by PMI for all $n \in \mathbb{N}$, both (B) and (D) qualify.

Let's assume there is no error and only one answer is correct. Sometimes, questions differentiate between proofs that can be done by simple inspection (like D for $n>1$) vs. those that require more manipulation in the inductive step (like B). However, the question just asks which can be proven by PMI.

Let's consider the possibility that the set $\mathbb{N}$ might mean something other than $\{1, 2, 3, ...\}$ in the context of the source, but the options imply positive integers starting from 1 (as base case $n=1$ is commonly used). Also, the question doesn't specify standard PMI starting at $n=1$, but options (A) and (C) fail at small positive integers, making them false statements for "all $n \in \mathbb{N}$".

Given that both (B) and (D) are true for all $n \in \mathbb{N}$ and provable by PMI, there is an ambiguity. However, if forced to choose one, statement (B) ($n^3 + 2n$ divisible by 3) is a very classic example used to illustrate the inductive step requiring algebraic manipulation and factoring, whereas (D) ($n!$ divisible by $n$) is almost definitionally true for $n>1$, and the base case handles $n=1$. Often, problems like (B) are presented specifically to demonstrate the inductive step. This is a weak basis for selection, but in a flawed question, it's sometimes the implicit intent.

Let's assume (B) is the intended answer, as it is a more standard example demonstrating algebraic manipulation in the inductive step for divisibility proofs.

Assuming the question intends to ask which statement is a standard example solvable by PMI with typical inductive step manipulation, the answer is likely (B).

However, mathematically, (D) is also provable by PMI for all $n \in \mathbb{N}$.

Let's proceed with the assumption that (B) is the intended answer due to it being a more representative example of inductive proof steps for divisibility.

The correct option is likely (B) $n^3 + 2n$ is divisible by 3 for all $n \in \mathbb{N}$.

Let's analyze the Assertion (A) and the Reason (R).

Assertion (A): If $P(1)$ is true and $P(k)$ implies $P(k+1)$ for some specific integer $k$, then $P(n)$ is true for all $n \geq 1$.

The Principle of Mathematical Induction requires that the implication $P(k) \implies P(k+1)$ holds for *all* integers $k$ from the base case onwards. The statement that it holds for "some specific integer $k$" is not sufficient to conclude that $P(n)$ is true for all $n \geq 1$. For example, $P(n): n=1$. $P(1)$ is true. $P(k) \implies P(k+1)$ would be "$k=1$ implies $k+1=1$". If we choose $k=1$, this becomes "$1=1$ implies $2=1$", which is false. The implication does not hold for $k=1$. If we chose a statement where $P(k) \implies P(k+1)$ *happened* to be true for *one* specific $k$, it still would not prove the statement for all $n$. The requirement is for *all* $k \geq 1$. Therefore, Assertion (A) is false.

Reason (R): The inductive step must hold for *all* positive integers $k \geq \text{base case value}$.

This is a correct statement of the requirement for the inductive step in the Principle of Mathematical Induction. If the base case is $P(n_0)$, then the inductive step requires proving $P(k) \implies P(k+1)$ for all integers $k \geq n_0$. In the context of Assertion (A) where the base case is $P(1)$, the inductive step must hold for all positive integers $k \geq 1$. Reason (R) accurately describes this requirement. Therefore, Reason (R) is true.

Assertion (A) is false, and Reason (R) is true. Reason (R) correctly states the condition that Assertion (A) fails to meet, explaining why Assertion (A) is false.

Based on the truth values of A and R, and the fact that R correctly describes the condition that A violates:

• A is false.
• R is true.

This matches option (D).

The correct option is (D) A is false but R is true.

The statement is $P(n): 1^2 + 2^2 + ... + n^2 = \frac{n(n+1)(2n+1)}{6}$.

The left-hand side (LHS) of $P(n)$ is the sum of the first $n$ squares: $1^2 + 2^2 + ... + n^2$.

The statement $P(k)$ is obtained by replacing $n$ with $k$: $1^2 + 2^2 + ... + k^2 = \frac{k(k+1)(2k+1)}{6}$. The LHS of $P(k)$ is the sum of the first $k$ squares: $1^2 + 2^2 + ... + k^2$.

The statement $P(k+1)$ is obtained by replacing $n$ with $k+1$. The LHS of $P(k+1)$ is the sum of the first $k+1$ squares:

LHS of $P(k+1) = 1^2 + 2^2 + ... + k^2 + (k+1)^2$

This sum consists of the sum of the first $k$ squares plus the $(k+1)$-th term squared.

LHS of $P(k+1) = \underbrace{(1^2 + 2^2 + ... + k^2)}_{\text{LHS of } P(k)} + (k+1)^2$

From the inductive hypothesis $P(k)$, we assume that the LHS of $P(k)$ is equal to the RHS of $P(k)$, i.e., $1^2 + 2^2 + ... + k^2 = \frac{k(k+1)(2k+1)}{6}$.

So, we can substitute the expression $\frac{k(k+1)(2k+1)}{6}$ for the sum of the first $k$ squares in the expression for the LHS of $P(k+1)$:

LHS of $P(k+1) = \frac{k(k+1)(2k+1)}{6} + (k+1)^2$

Let's look at the options:

(A) $\frac{k(k+1)(2k+1)}{6} + (k+1)^2$: This expression is obtained by substituting the assumed value of the sum of the first $k$ squares into the expression for the sum of the first $k+1$ squares. This is a correct expression for the LHS of $P(k+1)$ *when using the inductive hypothesis*.

(B) $1^2 + 2^2 + ... + k^2 + (k+1)^2$: This is the definition of the left-hand side of the statement $P(k+1)$ as the sum of the first $k+1$ squares.

(C) $\frac{(k+1)(k+2)(2(k+1)+1)}{6}$: This is the right-hand side of the statement $P(k+1)$, obtained by replacing $n$ with $k+1$ in the formula $\frac{n(n+1)(2n+1)}{6}$. It is NOT the left-hand side.

(D) Both (A) and (B) are correct expressions for the LHS of $P(k+1)$ based on $P(k)$. Expression (B) is the definition of the LHS of $P(k+1)$. Expression (A) is the LHS of $P(k+1)$ rewritten by substituting the assumed value from the inductive hypothesis for the sum of the first $k$ terms. In the context of the inductive step, we often start with the expression in (B), then rewrite the sum of the first $k$ terms using the inductive hypothesis to get the expression in (A), and then manipulate (A) to show it equals the RHS of $P(k+1)$. Both (A) and (B) are indeed correct expressions that represent the left-hand side of $P(k+1)$ in the context of the proof, where (A) is derived from (B) using the hypothesis $P(k)$.

Option (B) is the direct definition of the LHS of $P(k+1)$. Option (A) is what you get when you apply the inductive hypothesis to the LHS of $P(k+1)$. In the inductive step, you work with the LHS of $P(k+1)$, which is expressed as in (B), and then you use the inductive hypothesis, which leads to the expression in (A). Both are valid ways to represent or work with the LHS of $P(k+1)$ during the proof.

The correct option is (D) Both (A) and (B) are correct expressions for the LHS of $P(k+1)$ based on $P(k)$.

The Principle of Mathematical Induction (PMI) is a proof technique primarily used to prove statements about integers, most commonly positive integers.

Let's consider the types of statements typically proven using PMI:

(A) Formulae for sums of series (e.g., $1+2+...+n = \frac{n(n+1)}{2}$, $1^2+2^2+...+n^2 = \frac{n(n+1)(2n+1)}{6}$). These are very common applications of PMI.

(B) Divisibility properties of expressions involving $n$ (e.g., $n^3+2n$ is divisible by 3, $7^n - 3^n$ is divisible by 4). These are also common applications of PMI.

(C) Inequalities involving $n$ (e.g., $2^n > n$, $n! > 2^n$, Bernoulli's Inequality $(1+x)^n \geq 1+nx$). These are frequently proven using PMI.

(D) Existence of a solution to an equation (e.g., proving that the equation $x^2 - 4 = 0$ has a real solution). Proving the existence of a solution for a specific equation, or for a class of equations that don't depend on an integer $n$ in a way that builds sequentially, is generally not done using PMI. While induction might be used in some proofs related to the properties of functions or sets where existence is shown iteratively for structures indexed by integers, the core principle of proving the existence of a solution to a typical algebraic or transcendental equation is not the general domain of PMI.

PMI is most suitable for proving statements that establish a property holds for an entire sequence of integers by showing it holds for the first element and propagates to the next.

Proving the existence of a solution to an equation is usually done using other methods, such as algebraic manipulation, calculus (e.g., Intermediate Value Theorem), graphical analysis, or specific theorems related to the type of equation.

Therefore, the statement that is NOT generally proven using PMI is the existence of a solution to an equation.

The correct option is (D) Existence of a solution to an equation.

The statement is $P(n): 3^{2n} - 1$ is divisible by 8 for all positive integers $n$.

In the inductive step, we assume $P(k)$ is true for some positive integer $k$. The inductive hypothesis is $3^{2k} - 1 = 8m$ for some integer $m$. From this, we can write $3^{2k} = 8m + 1$.

The goal of the inductive step is to prove $P(k+1)$ is true. The statement $P(k+1)$ is obtained by replacing $n$ with $k+1$ in $P(n)$: $3^{2(k+1)} - 1$ is divisible by 8.

Let's analyze the options:

(A) $3^{2(k+1)} - 1 = 8 \cdot (\text{an integer})$: This is the definition of $3^{2(k+1)} - 1$ being divisible by 8. To prove $P(k+1)$ is true, we must show that the expression $3^{2(k+1)} - 1$ can be written in this form. So, this is the ultimate goal of the algebraic manipulation in the inductive step.

(B) $3^{2k+2} - 1$ can be written as $3^{2k} \cdot 3^2 - 1 = 9 \cdot 3^{2k} - 1 = 9(8m+1) - 1 = 72m + 9 - 1 = 72m + 8 = 8(9m+1)$: This option shows the algebraic steps involved in manipulating the expression $3^{2(k+1)} - 1$ and using the inductive hypothesis ($3^{2k} = 8m+1$) to show that the result is $8(9m+1)$. Since $m$ is an integer, $9m+1$ is an integer, so $3^{2(k+1)} - 1$ is indeed $8 \times (\text{an integer})$. These steps demonstrate that the expression is divisible by 8. This calculation is a valid way to show the goal in (A) is achieved.

(C) $P(k+1)$ is true: This is the statement that we are trying to prove in the inductive step. Proving $P(k+1)$ is true is the objective. The way to prove it is by showing that $3^{2(k+1)} - 1$ is divisible by 8, as stated in $P(k+1)$.

(D) All of the above: Option (C) states the overall objective of the inductive step. Option (A) states the mathematical condition that needs to be shown about the expression $3^{2(k+1)} - 1$ to prove $P(k+1)$. Option (B) provides the detailed algebraic calculation that successfully shows the condition in (A) is met, thereby proving (C) is true. All three options relate to what must be shown or done in the inductive step.

To prove $P(k+1)$ is true (C), we must show that $3^{2(k+1)} - 1$ is divisible by 8 (A). The calculation shown in (B) is a valid method to achieve this, by using the inductive hypothesis and algebraic manipulation to express $3^{2(k+1)} - 1$ in the form $8 \times (\text{integer})$. Therefore, all three options describe aspects of what must be shown or the process involved in showing it, within the inductive step.

The correct option is (D) All of the above.

A mathematical statement (or proposition) is a declarative sentence that is either true or false, but not both.

Let's examine each option:

(A) "The equation $x^2 = -1$ has no real solutions."

This is a declarative sentence. It makes a claim about the solutions of a specific equation within the domain of real numbers. In the real number system, the square of any real number is non-negative, so it cannot equal -1. Thus, the equation $x^2 = -1$ has no real solutions. This statement is true. Since it is a declarative sentence that is true, it is a valid mathematical statement.

(B) "All elephants are blue."

This is a declarative sentence. It makes a claim about a property of elephants. This statement is false (elephants are typically gray or brown). Since it is a declarative sentence that is false, it is a valid statement (though not a mathematical statement in the sense of being about mathematical objects, it is a declarative sentence with a definite truth value). However, in the context of a question asking for *mathematical* statements, options about real-world objects are generally excluded unless they are framed within a mathematical context (like probability or set theory, which isn't the case here). Assuming the question implies statements within pure mathematics, this might be excluded.

(C) "$17$ is a prime number."

This is a declarative sentence. It makes a claim about the number 17 and its property of being prime (a natural number greater than 1 that has no positive divisors other than 1 and itself). 17 is divisible only by 1 and 17. Thus, 17 is a prime number. This statement is true. Since it is a declarative sentence that is true and is about a mathematical concept (numbers and primality), it is a valid mathematical statement.

(D) "$y > x+5$, where $x$ and $y$ are real numbers."

This is an open inequality involving variables $x$ and $y$. Its truth value depends on the specific values of $x$ and $y$. For example, if $x=1$ and $y=10$, $10 > 1+5 \implies 10 > 6$, which is true. If $x=1$ and $y=5$, $5 > 1+5 \implies 5 > 6$, which is false. Since its truth value is not fixed but depends on the values of the variables, this is not a statement in itself. It is a propositional function or an open sentence.

Based on the definition of a mathematical statement as a declarative sentence that is unambiguously true or false:

• (A) is a true declarative sentence about a mathematical property. It is a valid mathematical statement.
• (B) is a false declarative sentence about a real-world property. In the context of "mathematical statements", this is usually excluded.
• (C) is a true declarative sentence about a mathematical property. It is a valid mathematical statement.
• (D) is an open sentence whose truth depends on variables. It is not a statement.

Therefore, options (A) and (C) are valid mathematical statements. Option (B) is a valid statement in general (has a truth value) but typically not considered a *mathematical* statement in this type of context unless part of a larger framework. Option (D) is not a statement.

Assuming the question asks for statements that are unambiguously true or false within a mathematical context, options (A) and (C) fit this definition.

The correct options are (A) The equation $x^2 = -1$ has no real solutions. and (C) $17$ is a prime number.

This question describes the conditions of the Principle of Mathematical Induction (PMI) when the base case is not necessarily $n=1$.

The given conditions are:

1. The base case: $P(3)$ is true.

2. The inductive step: $P(k)$ implies $P(k+1)$ for all integers $k \geq 3$.

According to the Principle of Mathematical Induction, if a statement $P(n)$ is true for a starting integer $n_0$ (the base case), and if the truth of $P(k)$ for any integer $k \geq n_0$ implies the truth of $P(k+1)$, then $P(n)$ is true for all integers $n \geq n_0$.

In this problem, the starting integer (base case) is $n_0 = 3$. The inductive step holds for all $k \geq 3$.

Applying PMI with the base case at $n=3$ and the inductive step holding for all $k \geq 3$:

• Since $P(3)$ is true, and the implication $P(3) \implies P(4)$ holds (because the inductive step holds for all $k \geq 3$, specifically for $k=3$), we can conclude that $P(4)$ is true.
• Since $P(4)$ is true (from the previous step), and the implication $P(4) \implies P(5)$ holds (because the inductive step holds for $k=4$), we can conclude that $P(5)$ is true.
• This process continues indefinitely.

Thus, the statement $P(n)$ is true for $n=3$, and for every integer greater than 3. This covers the set of integers $\{3, 4, 5, 6, ...\}$.

Let's look at the options:

(A) All positive integers $n$: Positive integers are $\{1, 2, 3, 4, ...\}$. The conditions only guarantee truth for $n \geq 3$, not necessarily for $n=1$ or $n=2$.

(B) All integers $n \geq 3$: This set is $\{3, 4, 5, 6, ...\}$. The conditions directly imply that $P(n)$ is true for all integers in this set.

(C) $n=3$ only: The conditions show that $P(n)$ is true for $n=3$ and also imply its truth for all integers greater than 3.

(D) $n=3, 4, 5, ...$ up to some limit: The Principle of Mathematical Induction proves the statement for the entire infinite sequence of integers starting from the base case, with no upper limit imposed by the principle itself.

The conditions given in the problem statement, according to the Principle of Mathematical Induction, prove the statement $P(n)$ for all integers $n$ greater than or equal to 3.

The correct option is (B) All integers $n \geq 3$.

The statement is $P(n): n^3 + n$ is divisible by 6 for all positive integers $n$.

The base case for proving this statement for all positive integers starting from $n=1$ is $P(1)$.

We check $P(1): 1^3 + 1 = 1 + 1 = 2$.

The statement $P(1)$ claims that 2 is divisible by 6. This is false, since $2 \div 6$ is not an integer.

If the base case $P(1)$ is false, then the statement "$P(n)$ is true for all positive integers $n$" is itself false, because it is false for $n=1$. The Principle of Mathematical Induction can only be used to prove statements that are actually true for the entire range of integers considered. If the base case is false, the principle cannot be applied to conclude that the statement is true for all integers in the domain starting from that base case.

Let's evaluate the options based on this understanding:

(A) The statement is false for $n=1$, so PMI cannot be applied starting from $n=1$. This is correct. PMI requires the base case to be true to establish the starting point of the inductive chain. If $P(1)$ is false, we cannot use PMI starting at $n=1$ to prove the statement for all $n \geq 1$.

(B) There must be an error in the problem statement or our understanding. The problem statement says "for all positive integers $n$". We found it's false for $n=1$. This means the statement as written ("... for all positive integers $n$") is actually false. There isn't necessarily an "error" in the statement itself, but the claim it makes is false. Our understanding is correct that $P(1)$ is false.

(C) PMI is not applicable to divisibility problems. This is incorrect. As shown in previous questions, PMI is a valid and common technique for proving divisibility statements for integers.

(D) We need to check a different base case. If the statement was true for $n \geq m$ for some $m > 1$, we could try to prove it for $n \geq m$ starting with base case $P(m)$. Let's check a few more values: $n=2: 2^3+2 = 8+2=10$ (not divisible by 6). $n=3: 3^3+3 = 27+3=30$. $30 \div 6 = 5$. $P(3)$ is true. $n=4: 4^3+4 = 64+4=68$ (not divisible by 6). $n=5: 5^3+5 = 125+5=130$ (not divisible by 6). $n=6: 6^3+6 = 216+6=222$. $222 \div 6 = 37$. $P(6)$ is true.

The statement $n^3+n$ is actually not divisible by 6 for all positive integers $n$ (it fails for $n=1, 2, 4, 5, ...$). It is true that $n^3+n = n(n^2+1) = n(n^2-1+2) = n((n-1)(n+1)+2) = n(n-1)(n+1) + 2n$. The term $n(n-1)(n+1)$ is the product of three consecutive integers, which is always divisible by 6. So $n^3+n = (\text{multiple of 6}) + 2n$. This sum is divisible by 6 if and only if $2n$ is divisible by 6, which means $n$ must be divisible by 3. So, $n^3+n$ is divisible by 6 if and only if $n$ is a multiple of 3. The statement is only true for $n=3, 6, 9, ...$, not for all positive integers.

Since the statement "$n^3+n$ is divisible by 6 for all positive integers $n$" is false, we cannot prove it using PMI starting from any base case for the claim "for all positive integers $n$". Option (D) suggests checking a different base case, which might be relevant if the statement was true for all integers from some $m > 1$ onwards. But the statement as a whole is false for the claimed domain (all positive integers).

Option (A) accurately reflects the immediate consequence of finding the base case $P(1)$ false when the statement is claimed to be true for all positive integers $n \geq 1$. It means the premise of the induction (that the statement is true for all $n \geq 1$) is incorrect, starting with the base case itself.

The most direct and correct implication of finding the base case $P(1)$ false for a statement claimed to be true for all $n \geq 1$ is that the statement is false for $n=1$, and therefore the claim for all $n \geq 1$ is false, making standard PMI starting from $n=1$ inapplicable to prove the *entire* claim.

The correct option is (A) The statement is false for $n=1$, so PMI cannot be applied starting from $n=1$.

The statement is $P(n): 1 \cdot 2 + 2 \cdot 3 + ... + n(n+1) = \frac{n(n+1)(n+2)}{3}$.

In the inductive step, we assume $P(k)$ is true for some positive integer $k$.

Inductive Hypothesis: $P(k)$ is true, which is $1 \cdot 2 + 2 \cdot 3 + ... + k(k+1) = \frac{k(k+1)(k+2)}{3}$.

We need to prove $P(k+1)$ is true. The statement $P(k+1)$ is obtained by replacing $n$ with $k+1$ in the statement $P(n)$.

LHS of $P(k+1) = 1 \cdot 2 + 2 \cdot 3 + ... + k(k+1) + (k+1)((k+1)+1) = 1 \cdot 2 + 2 \cdot 3 + ... + k(k+1) + (k+1)(k+2)$.

RHS of $P(k+1) = \frac{(k+1)((k+1)+1)((k+1)+2)}{3} = \frac{(k+1)(k+2)(k+3)}{3}$.

So, $P(k+1)$ is the statement: $1 \cdot 2 + 2 \cdot 3 + ... + k(k+1) + (k+1)(k+2) = \frac{(k+1)(k+2)(k+3)}{3}$.

To prove $P(k+1)$, we typically start with the LHS of $P(k+1)$ and use the inductive hypothesis $P(k)$.

LHS of $P(k+1) = \underbrace{(1 \cdot 2 + 2 \cdot 3 + ... + k(k+1))}_{\text{LHS of } P(k)} + (k+1)(k+2)$

By the inductive hypothesis, the sum of the first $k$ terms is equal to $\frac{k(k+1)(k+2)}{3}$.

So, LHS of $P(k+1) = \frac{k(k+1)(k+2)}{3} + (k+1)(k+2)$.

The goal is to show that this expression is equal to the RHS of $P(k+1)$, which is $\frac{(k+1)(k+2)(k+3)}{3}$.

So, proving $P(k+1)$ requires showing: $\frac{k(k+1)(k+2)}{3} + (k+1)(k+2) = \frac{(k+1)(k+2)(k+3)}{3}$.

Let's evaluate the options:

(A) $\frac{k(k+1)(k+2)}{3} + (k+1)(k+2) = \frac{(k+1)(k+2)(k+3)}{3}$: This is the equation that needs to be proven in the inductive step by substituting the inductive hypothesis into the LHS of $P(k+1)$ and setting it equal to the RHS of $P(k+1)$.

(B) $1 \cdot 2 + ... + k(k+1) + (k+1)(k+2) = \frac{(k+1)(k+2)(k+3)}{3}$: This is the statement $P(k+1)$ itself, which needs to be proven.

(C) Both (A) and (B) are equivalent representations of proving $P(k+1)$: Statement (B) is $P(k+1)$. Statement (A) is obtained from (B) by replacing the sum of the first $k$ terms with its assumed equal value from the inductive hypothesis $P(k)$. Proving (B) is equivalent to showing that the LHS of (B) equals the RHS of (B). Using the inductive hypothesis, the LHS of (B) is transformed into the expression on the left side of the equality in (A). Therefore, proving the equality in (A) is indeed equivalent to proving the statement $P(k+1)$ (B), under the assumption of $P(k)$. So, both are equivalent representations of the task in the inductive step.

(D) $\frac{k(k+1)(k+2)}{3} + k(k+1) = \frac{(k+1)(k+2)(k+3)}{3}$: This is incorrect. The term added to the sum of the first $k$ terms is the $(k+1)$-th term, which is $(k+1)(k+2)$, not $k(k+1)$.

Option (B) is the statement $P(k+1)$ itself. Option (A) is the form of the equation you need to verify algebraically in the inductive step, by utilizing the inductive hypothesis. Proving (A) is the practical task performed to show that (B) is true. Therefore, they are equivalent representations of the requirement for proving $P(k+1)$ in the inductive step.

The correct option is (C) Both (A) and (B) are equivalent representations of proving $P(k+1)$.

The Principle of Mathematical Induction involves two main parts:

1. The Base Case (or Basis Step): Proving the statement $P(n)$ is true for the initial value of $n$ (e.g., $n=1$).

2. The Inductive Step: Proving that if the statement $P(k)$ is true for some arbitrary integer $k$ (greater than or equal to the base case value), then the statement $P(k+1)$ is also true.

Within the inductive step, the assumption that $P(k)$ is true for some integer $k$ is a crucial part of the argument. This assumption is given a specific name.

Let's look at the options:

(A) Base Case: This is the first step of the induction, proving $P(n_0)$ is true, not the assumption about $P(k)$ in the inductive step.

(B) Inductive Step: This refers to the entire second part of the induction proof process (assuming $P(k)$ and proving $P(k+1)$), not specifically the assumption $P(k)$.

(C) Inductive Hypothesis: This is the standard term for the assumption made in the inductive step, which is that the statement $P(k)$ is true for some $k$ in the relevant range.

(D) Conclusion: This is the final statement of the proof, asserting that $P(n)$ is true for all $n$ in the specified range, based on the successful completion of the base case and inductive step.

The assumption that $P(k)$ is true in the inductive step is called the Inductive Hypothesis.

The correct option is (C) Inductive Hypothesis.

The Principle of Mathematical Induction (PMI) is used to prove statements about integers, typically starting from a specific integer and holding for all subsequent integers.

Let's examine the options:

(A) Proving that the sum of interior angles of a polygon with $n$ sides is $(n-2) \times 180^\circ$ for $n \geq 3$. A polygon with $n$ sides is defined for integers $n \geq 3$. This is a statement $P(n)$ about an integer $n$ (the number of sides) for a range of integers starting from 3. This type of statement, which relates a property to an integer index (number of sides), is a classic application of PMI. The base case would be a polygon with the smallest number of sides, a triangle ($n=3$), and we would show the formula holds. The inductive step would involve showing that if the formula holds for a polygon with $k$ sides, it holds for a polygon with $k+1$ sides. This is a valid application of PMI.

(B) Proving that a sequence is arithmetic. Proving that a specific sequence is arithmetic involves showing that the difference between consecutive terms is constant. This doesn't typically involve a proof by induction on an index $n$ to show a general property for all $n$, unless proving a property *about* arithmetic sequences for all $n$. For example, proving the formula for the $n$-th term or the sum of the first $n$ terms of an arithmetic sequence *can* be done by induction, but proving a sequence is arithmetic itself is usually done by checking the difference between terms.

(C) Proving that a function is continuous. Continuity is a property of functions defined on intervals of real numbers. Proofs of continuity involve limits and epsilon-delta arguments, not typically PMI on an integer index, unless proving a property *about* sequences of functions indexed by integers, for instance.

(D) Proving that a triangle is right-angled. Proving a specific triangle is right-angled involves using properties of angles or sides (e.g., Pythagorean theorem). This is a proof about a single geometric object, not a statement about a property that varies with an integer index $n$ for a series of objects.

Option (A) presents a statement about a property (sum of interior angles) that is indexed by an integer $n$ (number of sides) over a range of integers ($n \geq 3$). This structure is perfectly suited for proof by mathematical induction.

The correct option is (A) Proving that the sum of interior angles of a polygon with $n$ sides is $(n-2) \times 180^\circ$ for $n \geq 3$.

The statement is $P(n): n! > 2^n$ for $n \geq 4$.

In the inductive step, we assume $P(k)$ is true for some integer $k \geq 4$.

Inductive Hypothesis: $k! > 2^k$.

We want to prove $P(k+1)$ is true, i.e., $(k+1)! > 2^{k+1}$.

We start with the left side of $P(k+1)$:

$(k+1)! = (k+1) \cdot k!$

Using the inductive hypothesis $k! > 2^k$, and since $k \geq 4$, $(k+1)$ is a positive number (specifically, $k+1 \geq 5$). Multiplying both sides of the inequality $k! > 2^k$ by $(k+1)$ preserves the inequality sign:

$(k+1) \cdot k! > (k+1) \cdot 2^k$

So, $(k+1)! > (k+1) \cdot 2^k$.

Now we need to compare this with the right side of $P(k+1)$, which is $2^{k+1}$.

$2^{k+1} = 2 \cdot 2^k$.

To prove $(k+1)! > 2^{k+1}$, based on the inequality $(k+1)! > (k+1) \cdot 2^k$, it is sufficient to show that $(k+1) \cdot 2^k \geq 2^{k+1}$ or, more strongly for a strict inequality, $(k+1) \cdot 2^k > 2^{k+1}$.

Let's consider the inequality we need to make the link:

$(k+1) \cdot 2^k > 2^{k+1}$

Substitute $2^{k+1} = 2 \cdot 2^k$:

$(k+1) \cdot 2^k > 2 \cdot 2^k$

Since $2^k$ is a positive number (for integer $k$), we can divide both sides by $2^k$ without changing the direction of the inequality:

$k+1 > 2$

So, for $(k+1) \cdot 2^k$ to be strictly greater than $2^{k+1}$, the inequality $k+1 > 2$ needs to be true for the value of $k$ in the inductive step. Since the inductive step assumes $k \geq 4$, it means $k+1 \geq 5$. If $k+1 \geq 5$, then $k+1 > 2$ is definitely true.

The inequality that needs to be true for $k+1$ in order for $(k+1) \cdot 2^k > 2^{k+1}$ is $k+1 > 2$.

Let's look at the options:

(A) $k+1 > 1$: This is true for all positive integers $k$, including $k \geq 4$. But it's not the inequality needed to show $(k+1) \cdot 2^k > 2^{k+1}$.

(B) $k+1 > 2$: This is the specific inequality that, when true, ensures $(k+1) \cdot 2^k > 2^{k+1}$. This is true for $k > 1$. Since our inductive step applies for $k \geq 4$, this condition is met.

(C) $k+1 > 0$: This is true for all integers $k \geq -1$. It is true for $k \geq 4$, but not the specific condition needed.

(D) $k+1 \geq 4$: This is true because $k \geq 4$. If $k+1 \geq 4$, then $k+1 > 2$ is also true. However, $k+1 > 2$ is the direct condition derived from comparing $(k+1) \cdot 2^k$ and $2 \cdot 2^k$.

The required inequality for $k+1$ to bridge $(k+1) \cdot 2^k$ to $2^{k+1}$ is $k+1 > 2$.

The correct option is (B) $k+1 > 2$.

The Principle of Mathematical Induction (PMI) proves a statement $P(n)$ for a specific range of integers, starting from a base case and extending to all integers greater than or equal to that base case, provided the inductive step holds for that range.

In this scenario, the base case is $P(1)$, and the inductive step $P(k) \implies P(k+1)$ is proven for all $k \geq 1$. This establishes the truth of $P(n)$ for the set of integers $\{1, 2, 3, 4, ...\}$.

The question asks if this guarantees $P(n)$ is true for $n=0$ (assuming $P(0)$ is a meaningful statement).

PMI works like a chain reaction or domino effect. Proving $P(1)$ makes the first domino fall. Proving $P(k) \implies P(k+1)$ for $k \geq 1$ means that if any domino from the first one onwards falls, the next one also falls. This guarantees that all dominoes from the first one onwards will fall.

However, this says nothing about dominoes that come *before* the first one, such as a hypothetical $P(0)$ or negative values of $n$. The truth of $P(1), P(2), P(3), ...$ does not, by itself, imply the truth of $P(0)$ or $P(-1)$, etc.

Let's consider the options:

(A) Yes, always: This is incorrect. PMI only guarantees the statement for the range starting from the base case.

(B) No, PMI starting at $n=1$ only proves for $n=1, 2, 3, ...$: This is correct. The conclusion of PMI with base case $P(1)$ and inductive step for $k \geq 1$ is precisely that $P(n)$ is true for all integers $n \geq 1$. It makes no claim about values of $n$ less than 1.

(C) Yes, but only if the statement is also true for negative integers: This is incorrect. The truth for negative integers is irrelevant to the inductive proof for $n \geq 1$ and does not somehow extend the conclusion of that proof to $n=0$.

(D) It depends on the specific statement $P(n)$: While some statements might happen to be true for $n=0$ even if proven for $n \geq 1$, the *guarantee* from the PMI proof starting at $n=1$ does not cover $n=0$. The truth for $n=0$ would need to be proven separately or by a different method (perhaps induction starting at a different base case if applicable). The PMI proof starting at $n=1$ *itself* provides no information about $P(0)$.

The inductive proof starting at $n=1$ covers the integers $1, 2, 3, ...$. It does not provide a guarantee about the truth of the statement for $n=0$.

The correct option is (B) No, PMI starting at $n=1$ only proves for $n=1, 2, 3, ...$.

The statement is $P(n): 1 \cdot 3 + 2 \cdot 4 + ... + n(n+2) = \frac{n(n+1)(2n+7)}{6}$.

We need to find the value of $P(2)$. This means substituting $n=2$ into the statement $P(n)$ and evaluating both sides of the equality.

Left-Hand Side (LHS) of $P(2)$:

The LHS is the sum $1 \cdot 3 + 2 \cdot 4 + ... + n(n+2)$ up to $n=2$. So it is the sum of the first 2 terms:

LHS = $1 \cdot (1+2) + 2 \cdot (2+2)$

LHS = $1 \cdot 3 + 2 \cdot 4$

LHS = $3 + 8$

LHS = $11$

This matches option (A), stating the value of the LHS.

Right-Hand Side (RHS) of $P(2)$:

The RHS is the formula $\frac{n(n+1)(2n+7)}{6}$ evaluated at $n=2$.

RHS = $\frac{2(2+1)(2(2)+7)}{6}$

RHS = $\frac{2(3)(4+7)}{6}$

RHS = $\frac{2 \cdot 3 \cdot 11}{6}$

RHS = $\frac{6 \cdot 11}{6}$

RHS = $11$

This matches option (B), stating the value of the RHS.

The statement $P(2)$ is the equality between the LHS and RHS when $n=2$. So, $P(2)$ is the statement $11 = 11$. Since $11$ is indeed equal to $11$, the statement $P(2)$ is true.

Let's look at the options:

(A) $1 \cdot 3 + 2 \cdot 4 = 3 + 8 = 11$ (LHS): This correctly calculates the value of the left-hand side of $P(2)$.

(B) $\frac{2(2+1)(2(2)+7)}{6} = \frac{2 \cdot 3 \cdot 11}{6} = 11$ (RHS): This correctly calculates the value of the right-hand side of $P(2)$.

(C) $11$ (LHS) and $11$ (RHS): This states the calculated values of both the LHS and RHS of $P(2)$. The statement $P(2)$ itself is the assertion that these two values are equal.

(D) All of the above demonstrate that $P(2)$ is true: Option (A) shows the value of the LHS. Option (B) shows the value of the RHS. Showing that LHS = RHS for $n=2$ demonstrates that $P(2)$ is true. Both (A) and (B) provide the necessary calculations to verify this equality, and (C) explicitly states the result of these calculations. Therefore, all options contribute to demonstrating or stating the truth of $P(2)$.

Options (A), (B), and (C) provide the necessary information and calculations to confirm that $P(2)$ is true. Option (D) correctly summarizes this. Calculating the LHS and RHS (A and B) and showing they are equal (implied by C) is the process of verifying $P(2)$, and the fact that they are equal demonstrates that $P(2)$ is true.

The correct option is (D) All of the above demonstrate that $P(2)$ is true.

The sequence is defined by $a_1 = 2$ and $a_{n+1} = 2a_n + 1$ for $n \geq 1$. The statement is $P(n): a_n = 3 \cdot 2^{n-1} - 1$.

We are given the first few terms of the sequence in a table:

We need to check for which values of $n$ from the table the statement $P(n): a_n = 3 \cdot 2^{n-1} - 1$ is true by comparing the values from the table with the values given by the formula.

• For $n=1$: According to the table, $a_1 = 2$. According to the formula, $a_1 = 3 \cdot 2^{1-1} - 1 = 3 \cdot 2^0 - 1 = 3 \cdot 1 - 1 = 3 - 1 = 2$. Since $2=2$, $P(1)$ is true.
• For $n=2$: According to the table, $a_2 = 5$. According to the formula, $a_2 = 3 \cdot 2^{2-1} - 1 = 3 \cdot 2^1 - 1 = 3 \cdot 2 - 1 = 6 - 1 = 5$. Since $5=5$, $P(2)$ is true.
• For $n=3$: According to the table, $a_3 = 11$. According to the formula, $a_3 = 3 \cdot 2^{3-1} - 1 = 3 \cdot 2^2 - 1 = 3 \cdot 4 - 1 = 12 - 1 = 11$. Since $11=11$, $P(3)$ is true.
• For $n=4$: According to the table, $a_4 = 23$. According to the formula, $a_4 = 3 \cdot 2^{4-1} - 1 = 3 \cdot 2^3 - 1 = 3 \cdot 8 - 1 = 24 - 1 = 23$. Since $23=23$, $P(4)$ is true.
• For $n=5$: According to the table, $a_5 = 47$. According to the formula, $a_5 = 3 \cdot 2^{5-1} - 1 = 3 \cdot 2^4 - 1 = 3 \cdot 16 - 1 = 48 - 1 = 47$. Since $47=47$, $P(5)$ is true.

Based on the values given in the table, the statement $P(n): a_n = 3 \cdot 2^{n-1} - 1$ is true for all the values of $n$ provided in the table, which are $n=1, 2, 3, 4, 5$.

The question asks for which values of $n$ *based on the table* is $P(n)$ true. The table provides values for $n=1, 2, 3, 4, 5$. Our verification shows $P(n)$ is true for all these values.

Let's look at the options:

(A) $n=1, 2, 3, 4, 5$: This matches the values of $n$ in the table for which we found $P(n)$ to be true.

(B) Only $n=1$: This is incorrect, as $P(n)$ is also true for $n=2, 3, 4, 5$ based on the table.

(C) Only $n=5$: This is incorrect.

(D) Cannot be determined from the table alone: This is incorrect. By using the values from the table for $a_n$ and the given formula, we can determine for which tabulated values of $n$ the statement is true.

The correct option is (A) $n=1, 2, 3, 4, 5$.

The statement is $P(n): a_n = 3 \cdot 2^{n-1} - 1$. The sequence is defined by $a_{n+1} = 2a_n + 1$. We are in the inductive step of proving $P(n)$ for $n \geq 1$.

We assume $P(k)$ is true for some positive integer $k \geq 1$.

Inductive Hypothesis: $a_k = 3 \cdot 2^{k-1} - 1$.

We need to prove $P(k+1)$ is true, which is the statement $a_{k+1} = 3 \cdot 2^{(k+1)-1} - 1 = 3 \cdot 2^k - 1$.

We start with the definition of $a_{k+1}$ from the recurrence relation:

Now we use the inductive hypothesis $a_k = 3 \cdot 2^{k-1} - 1$ and substitute the expression for $a_k$ into the recurrence relation:

$a_{k+1} = 2(\underbrace{3 \cdot 2^{k-1} - 1}_{a_k}) + 1$

Let's look at the options:

(A) $a_{k+1} = 2(3 \cdot 2^{k-1}) + 1$: This is incorrect. When substituting $a_k$, the entire expression for $a_k$ ($3 \cdot 2^{k-1} - 1$) must be multiplied by 2, not just the first term.

(B) $a_{k+1} = 2(3 \cdot 2^{k-1} - 1) + 1 = 6 \cdot 2^{k-1} - 2 + 1 = 3 \cdot 2 \cdot 2^{k-1} - 1 = 3 \cdot 2^k - 1$: This option correctly substitutes the inductive hypothesis into the recurrence relation and then performs the algebraic steps to simplify the expression. The steps shown are:

• Start with $a_{k+1} = 2a_k + 1$.
• Substitute $a_k = 3 \cdot 2^{k-1} - 1$: $a_{k+1} = 2(3 \cdot 2^{k-1} - 1) + 1$.
• Distribute the 2: $a_{k+1} = 2 \cdot (3 \cdot 2^{k-1}) - 2 \cdot 1 + 1 = 6 \cdot 2^{k-1} - 2 + 1$.
• Rewrite $6 \cdot 2^{k-1}$ as $(3 \cdot 2) \cdot 2^{k-1} = 3 \cdot (2 \cdot 2^{k-1}) = 3 \cdot 2^{k-1+1} = 3 \cdot 2^k$. Combine constants: $-2+1 = -1$.
• So, $a_{k+1} = 3 \cdot 2^k - 1$. This is the right side of the statement $P(k+1)$.

The calculation in option (B) shows that starting from $a_{k+1} = 2a_k + 1$ and using the inductive hypothesis, we arrive at $a_{k+1} = 3 \cdot 2^k - 1$, which is exactly the statement $P(k+1)$. This completes the inductive step.

(C) $a_{k+1} = 3 \cdot 2^k - 1$: This is the statement $P(k+1)$ that we need to prove is true for $a_{k+1}$. It's the goal expression, not the process of substitution and manipulation.

(D) The steps shown in option (B) complete the inductive step: This is true. The steps in option (B) correctly show that $P(k+1)$ follows from $P(k)$ using the recurrence relation.

The question asks what expression we should consider for the inductive step *using the recurrence relation $a_{k+1} = 2a_k + 1$, substitute the inductive hypothesis*. Option (B) shows this exact process of substitution and the subsequent simplification that proves $P(k+1)$. Options (A) and (C) are parts of the process or incorrect applications. Option (D) describes the outcome of option (B).

Option (B) provides the actual calculation and derivation that is performed in the inductive step using the recurrence relation and the inductive hypothesis to prove $P(k+1)$.

The correct option is (B) $a_{k+1} = 2(3 \cdot 2^{k-1} - 1) + 1 = 6 \cdot 2^{k-1} - 2 + 1 = 3 \cdot 2 \cdot 2^{k-1} - 1 = 3 \cdot 2^k - 1$.

The Principle of Mathematical Induction (PMI) is a fundamental axiom or theorem about the structure of the set of natural numbers (or positive integers, depending on the definition). It is one of the key properties that characterizes these sets.

Let's analyze the options:

(A) PMI can prove statements about any set of numbers: This is false. PMI is specifically applicable to sets that have an ordered, sequential structure starting from a first element, like integers, not arbitrary sets like real numbers or complex numbers in general.

(B) PMI is a property that holds true for the set of positive integers (natural numbers) and is used to prove statements *about* these numbers: This is a correct description. PMI itself is a property of the set of positive integers (often stated as an axiom, like the Peano axioms, or proven from other axioms). It allows us to prove statements that claim a property holds for all positive integers by verifying the base case and the step-by-step implication.

(C) The set of positive integers is defined using PMI: This is partially true in some foundational systems of mathematics (like the Peano axioms, where the Induction Axiom is one of the axioms defining natural numbers). However, saying "defined using PMI" is a stronger claim than "PMI is a property that holds true for the set". While related to the definition, PMI is more accurately described as a property or axiom that governs how proofs work within that set.

(D) There is no direct relationship: This is false. There is a very direct and fundamental relationship between PMI and the set of positive integers.

Option (B) most accurately describes the relationship from the perspective of using PMI as a proof technique. PMI is a principle that holds for the set of positive integers due to their structure, and this principle is then used as a tool to prove various properties or statements that apply to all positive integers.

While (C) touches upon foundational aspects (Peano axioms), (B) is a better description of the practical relationship encountered when using PMI in standard mathematical proofs about positive integers.

The correct option is (B) PMI is a property that holds true for the set of positive integers (natural numbers) and is used to prove statements *about* these numbers.

The Principle of Mathematical Induction (PMI) is a proof technique used for statements that are claimed to be true for an infinite sequence of integers, starting from a base case. The statement must typically involve an integer variable $n$ and assert a property $P(n)$.

Let's review the types of statements and their suitability for PMI:

(A) Statements involving sums or products of terms depending on $n$: For example, proving formulas for $\sum_{i=1}^n f(i)$ or $\prod_{i=1}^n f(i)$. These are classic applications of PMI.

(B) Statements about inequalities involving $n$: For example, proving $2^n > n$ or $n! > 2^n$. These are common applications of PMI.

(C) Statements about properties of geometric figures: For example, proving that the sum of angles in a triangle is $180^\circ$, or that a specific quadrilateral is a rectangle. Statements about individual geometric figures (like a specific triangle or a specific circle) do not typically involve an integer variable $n$ in a way suitable for PMI. While properties of polygons with $n$ sides (where $n$ is the number of sides) can be proven by induction (as seen in a previous question), general statements about the properties of geometric figures that are not parameterized by an integer index $n$ are not proven by PMI.

(D) Statements about divisibility of expressions involving $n$: For example, proving $n^3+2n$ is divisible by 3. These are common applications of PMI.

Statements about specific geometric figures or general properties of geometric figures that do not depend on an integer index $n$ in a recursive manner are typically proven using axioms and theorems of geometry, not PMI.

Therefore, statements about properties of geometric figures (unless structured to depend on an integer $n$, like polygons with $n$ sides) are typically NOT proven using PMI.

The correct option is (C) Statements about properties of geometric figures.

The statement is $P(n): n^3 + 5n$ is divisible by 6 for all positive integers $n$. We need to identify the true statements about verifying this using PMI.

Let's examine each option:

(A) Base case $P(1)$: $1^3 + 5(1) = 1 + 5 = 6$. Is 6 divisible by 6? Yes. The calculation is correct, and the conclusion that $P(1)$ is true is correct. This is a true statement about verifying the base case $P(1)$.

(B) Base case $P(2)$: $2^3 + 5(2) = 8 + 10 = 18$. Is 18 divisible by 6? Yes, $18 \div 6 = 3$. The calculation is correct, and the conclusion that $P(2)$ is true is correct. While $P(1)$ is the standard base case for proving for all $n \geq 1$, verifying $P(2)$ is a valid step in checking the statement's truth for particular values of $n$. This is a true statement about verifying $P(2)$.

(C) Inductive step: Assume $k^3 + 5k$ is divisible by 6. Consider $(k+1)^3 + 5(k+1)$.

The expansion and rearrangement shown are:

$(k+1)^3 + 5(k+1) = (k^3 + 3k^2 + 3k + 1) + (5k + 5)$ (correct expansion and distribution)

$= k^3 + 3k^2 + 8k + 6$ (correct combination of terms)

The statement in option (C) then rearranges this as $(k^3 + 5k) + 3k^2 + 3k + 6$. Let's check this rearrangement:

$(k^3 + 5k) + 3k^2 + 3k + 6 = k^3 + 5k + 3k^2 + 3k + 6 = k^3 + 3k^2 + 8k + 6$. This matches the expanded form. So the rearrangement is correct.

The statement then factors $3k^2 + 3k$ as $3k(k+1)$: $(k^3 + 5k) + 3k(k+1) + 6$. This factoring is correct.

So, option (C) correctly shows the algebraic manipulation to rewrite $(k+1)^3 + 5(k+1)$ in a form that includes the inductive hypothesis term $(k^3+5k)$ and other terms. This is a valid and necessary step in the inductive proof. This is a true statement about the inductive step.

(D) In the inductive step, since $(k^3 + 5k)$ is divisible by 6 (by hypothesis), $3k(k+1)$ is divisible by 6 (since $k(k+1)$ is always even, $3k(k+1)$ is divisible by $3 \times 2 = 6$), and 6 is divisible by 6, their sum is divisible by 6. Thus $P(k+1)$ is true.

This option provides the logical reasoning to complete the inductive step based on the expression derived in (C). It correctly states that:

• $(k^3 + 5k)$ is divisible by 6 due to the inductive hypothesis.
• $k(k+1)$ is the product of two consecutive integers, so it is always even. Thus, $k(k+1) = 2j$ for some integer $j$.
• $3k(k+1) = 3(2j) = 6j$. So, $3k(k+1)$ is always divisible by 6.
• 6 is divisible by 6.
• The sum $(k^3 + 5k) + 3k(k+1) + 6$ is the sum of three terms, each divisible by 6. The sum of numbers divisible by 6 is also divisible by 6.

Therefore, $(k+1)^3 + 5(k+1)$ is divisible by 6. This means $P(k+1)$ is true. This option correctly describes the final steps and reasoning to complete the inductive proof. This is a true statement about the inductive step.

All four options (A), (B), (C), and (D) present true statements related to verifying or performing the proof of $P(n)$ using PMI. (A) and (B) verify specific cases. (C) shows correct algebraic manipulation in the inductive step. (D) provides the correct logical deduction to complete the inductive step using the result from (C).

Since it is a "Multiple Correct Answer(s)" question, and we have identified (A), (B), (C), and (D) as true statements about verifying this using PMI, all of them are correct answers.

The correct options are (A), (B), (C), and (D).

The statement is $P(n): \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + ... + \frac{1}{n(n+1)} = \frac{n}{n+1}$.

The right-hand side (RHS) of the statement $P(n)$ is the expression $\frac{n}{n+1}$.

The statement $P(k+1)$ is obtained by replacing $n$ with $k+1$ in the statement $P(n)$.

So, the RHS of $P(k+1)$ is the expression $\frac{n}{n+1}$ evaluated at $n=k+1$.

RHS of $P(k+1) = \frac{(k+1)}{(k+1)+1}$

Simplifying the denominator:

RHS of $P(k+1) = \frac{k+1}{k+2}$

Let's look at the options:

(A) $\frac{k}{k+1} + \frac{1}{(k+1)(k+2)}$: This is the left-hand side of $P(k+1)$ expressed using the inductive hypothesis $P(k)$ (sum of first $k$ terms) plus the $(k+1)$-th term. This is not the right-hand side.

(B) $\frac{k+1}{k+2}$: This is the simplified form of the RHS of $P(k+1)$, obtained by substituting $n=k+1$ into the formula $\frac{n}{n+1}$.

(C) $\frac{k+1}{(k+1)+1}$: This is the expression for the RHS of $P(k+1)$ with $n$ replaced by $k+1$, before simplifying the denominator.

(D) Both (B) and (C) are correct: Option (C) is the expression for the RHS of $P(k+1)$ before simplification. Option (B) is the simplified form of the RHS of $P(k+1)$. Both represent the right-hand side of the statement $P(k+1)$.

The right-hand side of the statement $P(k+1)$ is obtained by substituting $n=k+1$ into the RHS expression of $P(n)$. This gives $\frac{k+1}{(k+1)+1}$, which simplifies to $\frac{k+1}{k+2}$. Both forms are correct representations of the RHS of $P(k+1)$.

The correct option is (D) Both (B) and (C) are correct.

The statement is $P(n): 2^n > n^3$. We need to find the minimum value of $n$ from the options for which the statement is true.

Let's check the truth of the statement for each option:

• For $n=1$: $P(1): 2^1 > 1^3 \implies 2 > 1$. This is True.
• For $n=5$: $P(5): 2^5 > 5^3 \implies 32 > 125$. This is False.
• For $n=9$: $P(9): 2^9 > 9^3 \implies 512 > 729$. This is False.
• For $n=10$: $P(10): 2^{10} > 10^3 \implies 1024 > 1000$. This is True.

The statement $2^n > n^3$ is true for $n=1$ and $n=10$ among the given options.

The minimum value of $n$ among the options for which the statement holds true is the smallest value from $\{1, 10\}$, which is 1.

Based on the literal wording of the question and the options provided, the minimum value of $n$ for which the statement holds true is 1.

The correct option is (A) 1.

The Principle of Mathematical Induction (PMI) is a proof technique used to prove statements about the properties of integers, specifically for a range of integers starting from a particular base value.

The principle relies on the well-ordered nature of the set of integers (or a subset of integers bounded below). The base case establishes the property for the first element, and the inductive step shows that if the property holds for an element, it holds for the next one in the sequence.

Let's consider the sets provided:

(A) Integers: This set includes positive integers, negative integers, and zero. Standard PMI proves statements for integers greater than or equal to a base case ($n \geq n_0$). It does not directly cover all integers (e.g., it doesn't prove it for negative integers unless specifically formulated). However, it proves for a subset of integers.

(B) Rational Numbers: Rational numbers are dense and do not have the sequential structure $n, n+1$ that PMI relies on.

(C) Natural Numbers (Positive Integers): Natural numbers (often starting from 1) are the most common and direct domain for standard PMI proofs. The set $\{1, 2, 3, ...\}$ has the perfect structure for PMI.

(D) Real Numbers: Real numbers form a continuum and do not have the sequential structure required for standard PMI.

While PMI can be extended or adapted to prove statements about sets that can be put into one-to-one correspondence with natural numbers, its core application and the set for which it is a fundamental principle is the set of natural numbers (positive integers).

The phrase "all $n$ in the set of ________" implies the domain of the proof. The most typical and direct application of PMI is to prove statements about all natural numbers (positive integers) from a certain point onwards.

The correct option is (C) Natural Numbers (Positive Integers).

The statement is $P(n): 2+4+6+...+2n = n(n+1)$. The left-hand side is the sum of the first $n$ even numbers.

We are asked for the sum of the first $k+1$ even numbers according to this pattern. This corresponds to the left-hand side of the statement $P(n)$ when $n$ is replaced by $k+1$.

The pattern of the sum is $2+4+6+...+2n$. The $n$-th term in this sum is $2n$.

The sum of the first $k+1$ even numbers is obtained by stopping the sum at the $(k+1)$-th term. The $(k+1)$-th term is found by replacing $n$ with $k+1$ in the expression for the $n$-th term ($2n$), which gives $2(k+1)$.

The sum of the first $k+1$ even numbers is $2+4+6+...+2k+2(k+1)$.

This sum can be written as the sum of the first $k$ even numbers plus the $(k+1)$-th even number.

Sum of first $k+1$ even numbers = $\underbrace{(2+4+6+...+2k)}_{\text{Sum of first } k \text{ even numbers}} + 2(k+1)$

Let's look at the options:

(A) $k(k+1)$: This is the formula for the sum of the first $k$ even numbers (the RHS of $P(k)$), not the sum of the first $k+1$ even numbers.

(B) $2(k+1)$: This is the $(k+1)$-th even number, not the sum of the first $k+1$ even numbers.

(C) $k(k+1) + 2(k+1)$: This expression correctly represents the sum of the first $k+1$ even numbers by taking the sum of the first $k$ terms (which is $k(k+1)$ according to the formula) and adding the $(k+1)$-th term ($2(k+1)$).

(D) $(k+1)(k+2)$: This is the expected formula for the sum of the first $k+1$ even numbers (obtained by replacing $n$ with $k+1$ in the RHS of $P(n)$). $P(k+1)$ states $2+4+6+...+2(k+1) = (k+1)((k+1)+1) = (k+1)(k+2)$. While this is the value the sum should equal, the question asks for the "sum of the first $k+1$ even numbers according to this pattern". The pattern is the sum $2+4+...+2n$. The sum according to this pattern up to $k+1$ terms is $2+4+...+2(k+1)$. This sum is also equal to the sum up to $k$ terms plus the $(k+1)$-th term.

The sum of the first $k+1$ even numbers is $2+4+...+2k+2(k+1)$. Using the pattern $P(n)$, the sum up to $k$ terms is $k(k+1)$. So the sum up to $k+1$ terms is $k(k+1) + 2(k+1)$.

Option (C) correctly expresses the sum of the first $k+1$ even numbers as the sum of the first $k$ terms (given by the formula in $P(k)$) plus the $(k+1)$-th term.

The correct option is (C) $k(k+1) + 2(k+1)$.

Let's analyze the Assertion (A) and the Reason (R).

Assertion (A): The statement "Every prime number is odd" can be disproven by a counterexample.

A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. The prime numbers start with 2, 3, 5, 7, 11, ...

An odd number is an integer that is not divisible by 2. The number 2 is a prime number, and 2 is an even number (it is divisible by 2). Therefore, 2 is a prime number that is not odd.

The number 2 serves as a counterexample to the universal statement "Every prime number is odd". Since a counterexample exists, the statement is false, and the existence of this counterexample allows the statement to be disproven.

Thus, Assertion (A) is true.

Reason (R): Mathematical induction is primarily used to prove statements that hold for all natural numbers, not to disprove statements or prove properties of specific numbers.

Mathematical induction is a method used to prove that a property holds for all integers in a specified range (typically natural numbers from a base case onwards). It is a proof technique for establishing the truth of certain universal statements over the integers.

To disprove a universal statement (a statement claiming something is true for *all* members of a set), one only needs to find a single instance where the statement is false. This single instance is called a counterexample. Mathematical induction is not used for finding counterexamples or disproving statements in general.

Furthermore, mathematical induction is used to prove properties that hold for an entire set or sequence of integers, not to prove properties of specific, individual numbers (e.g., proving that 7 is a prime number is done by checking its divisors, not by induction).

Thus, Reason (R) is a correct description of the purpose and limitations of Mathematical Induction. Reason (R) is true.

Now let's consider if Reason (R) is the correct explanation of Assertion (A).

Assertion (A) states that the specific statement "Every prime number is odd" can be disproven by a counterexample. Reason (R) explains that mathematical induction is a tool for *proving* statements for all natural numbers and is *not* used for disproving statements. This explains *why* one would use a counterexample to disprove the statement in (A) rather than attempting to use mathematical induction. Reason (R) clarifies the role of induction and, by contrast, supports the method described in Assertion (A) for disproving statements of this type.

Therefore, Reason (R) correctly explains why disproof by counterexample is the relevant technique here, by stating what PMI is and is not used for.

Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation for why a counterexample is used for disproving the statement in (A).

The correct option is (A) Both A and R are true and R is the correct explanation of A.

The Principle of Mathematical Induction (PMI) is used to prove that a statement $P(n)$ holds for all integers $n$ greater than or equal to some specific starting integer $n_0$. This starting integer $n_0$ is called the base case.

For PMI to be applicable, the set of integers for which the statement is being proven must be a set of the form $\{n_0, n_0+1, n_0+2, ...\}$, where $n_0$ is the smallest integer in the set. The base case is proving $P(n_0)$ is true.

Let's consider the options for potential base cases $n_0$:

• (A) $n=1$: This is a valid base case for proving statements about positive integers (the set $\{1, 2, 3, ...\}$). $n_0=1$.
• (B) $n=0$: This is a valid base case for proving statements about non-negative integers (the set $\{0, 1, 2, ...\}$), provided the statement $P(n)$ is defined for $n=0$. $n_0=0$.
• (D) $n=5$: This is a valid base case for proving statements about integers $n \geq 5$ (the set $\{5, 6, 7, ...\}$). $n_0=5$.

Now consider option (C) $n=-1$. Can $n=-1$ be a base case? Yes, if we are proving a statement $P(n)$ for all integers $n \geq -1$ (the set $\{-1, 0, 1, 2, ...\}$), and $P(n)$ is defined for these values. For example, we could prove by induction that the sum of integers from -1 to $n$ is $\frac{(n+1)(n-1+2)}{2} = \frac{(n+1)(n+1)}{2}$ (which simplifies to $\frac{(n+1)^2}{2}$? No, sum from $a$ to $n$ is $\frac{(n-a+1)(a+n)}{2}$ from -1 to $n$ is $\frac{(n-(-1)+1)(-1+n)}{2} = \frac{(n+2)(n-1)}{2}$?). Wait, let's use a simpler example. Consider the statement $P(n): n \geq -1$. Base case $n=-1$: $P(-1): -1 \geq -1$. True. Inductive step: Assume $P(k): k \geq -1$ is true for $k \geq -1$. Prove $P(k+1): k+1 \geq -1$. Since $k \geq -1$, adding 1 to both sides gives $k+1 \geq -1+1$, so $k+1 \geq 0$. Since $0 \geq -1$, $k+1 \geq 0 \geq -1$ is true. So $P(k+1)$ is true. Thus $P(n): n \geq -1$ is true for all $n \geq -1$ by PMI with base case $n=-1$. This shows $n=-1$ *can* be a base case.

However, the options represent specific integer values. In the standard context of introductory PMI, statements $P(n)$ are typically defined for $n \in \mathbb{N}$ (positive integers $\geq 1$) or $n \in \mathbb{N}_0$ (non-negative integers $\geq 0$). If the statement $P(n)$ implicitly refers to properties only defined for $n \geq 0$ or $n \geq 1$ (e.g., formulas involving sums from 0 or 1 to $n$, or properties of collections of $n$ objects), then a negative integer like $n=-1$ would not be a valid starting point because $P(-1)$ might not be defined or relevant in that domain. While technically PMI can start at any integer base $n_0$ for which $P(n)$ is defined for all $n \geq n_0$, option (C) represents a value that would fall outside the typical domain of application ($n \geq 0$ or $n \geq 1$) for many introductory problems.

Given that (A), (B), and (D) are all common base case values for typical domains ($n \geq 1$, $n \geq 0$, $n \geq 5$), option (C) $n=-1$ is the value that is least likely to be a valid base case in standard problems, often because the statement $P(n)$ is not defined or meaningful for negative $n$ in those contexts.

Assuming the question implies a typical context where statements are defined for $n \geq 0$, a base case of $n=-1$ would be invalid.

The correct option is (C) $n=-1$.

The statement is $P(n): 1^2 + 3^2 + ... + (2n-1)^2 = \frac{n(2n-1)(2n+1)}{3}$. The left-hand side (LHS) is the sum of the squares of the first $n$ odd positive integers.

The terms in the sum on the LHS are of the form $(2i-1)^2$ for $i=1, 2, ..., n$. The $n$-th term is $(2n-1)^2$.

The LHS of $P(n)$ is $1^2 + 3^2 + ... + (2n-1)^2$.

The statement $P(k)$ has LHS $1^2 + 3^2 + ... + (2k-1)^2$.

The statement $P(k+1)$ has LHS $1^2 + 3^2 + ... + (2(k+1)-1)^2$.

The terms in the sum for $P(k+1)$ are the terms in the sum for $P(k)$ plus one additional term. This additional term is the $(k+1)$-th term in the series.

The $n$-th term is $(2n-1)^2$. To find the $(k+1)$-th term, we substitute $n=k+1$ into the expression for the $n$-th term:

$(k+1)$-th term = $(2(k+1)-1)^2$

$= (2k + 2 - 1)^2$

$= (2k+1)^2$

So, the LHS of $P(k+1)$ is $(1^2 + 3^2 + ... + (2k-1)^2) + (2k+1)^2$.

The term added to the LHS of $P(k)$ to get the LHS of $P(k+1)$ is the $(k+1)$-th term, which is $(2k+1)^2$.

Let's look at the options:

(A) $(2(k+1)-1)^2 = (2k+1)^2$: This is the correct expression for the $(k+1)$-th term, which is the term added to the LHS of $P(k)$ to get the LHS of $P(k+1)$.

(B) $(2k-1)^2$: This is the $k$-th term of the series, not the term added when going from $P(k)$ to $P(k+1)$.

(C) $(k+1)^2$: This would be the $(k+1)$-th term if the series was $1^2 + 2^2 + 3^2 + ... + n^2$, but this series consists of squares of odd numbers.

(D) $\frac{(k+1)(2(k+1)-1)(2(k+1)+1)}{3}$: This is the right-hand side of the statement $P(k+1)$, not a term added to the left-hand side.

The term added to the LHS of $P(k)$ to obtain the LHS of $P(k+1)$ is the $(k+1)$-th term of the series, which is $(2(k+1)-1)^2 = (2k+1)^2$.

The correct option is (A) $(2(k+1)-1)^2 = (2k+1)^2$.

The statement is $P(n): n^2+n+41$ is a prime number. We are given that $P(1)$ is true ($1^2+1+41 = 43$, which is prime) and $P(2)$ is true ($2^2+2+41 = 4+2+41 = 47$, which is prime).

The Principle of Mathematical Induction requires two conditions to be met to prove a statement $P(n)$ for all positive integers $n \geq 1$:

1. The Base Case: $P(1)$ must be true.

2. The Inductive Step: For any positive integer $k \geq 1$, if $P(k)$ is true, then $P(k+1)$ must also be true ($P(k) \implies P(k+1)$).

Checking the statement for a few initial values (like $n=1$ and $n=2$) only verifies the base case (or a few cases) and potentially provides evidence for a conjecture. It does not constitute a proof that the statement holds for all positive integers. A statement holding for the first few values is not sufficient to conclude it holds for all values via PMI.

For example, $P(3): 3^2+3+41 = 9+3+41 = 53$ (prime). $P(4): 4^2+4+41 = 16+4+41 = 61$ (prime). This sequence $n^2+n+41$ famously generates prime numbers for $n=1, 2, ..., 39$. However, for $n=40$, $P(40): 40^2+40+41 = 1600+40+41 = 1681 = 41^2$. Since 1681 is divisible by 41 (besides 1 and 1681), it is not prime. So, $P(40)$ is false.

This demonstrates that even if a pattern holds for many initial values, it might eventually fail.

Let's look at the options:

(A) Yes, because the pattern holds for the first two values. This is incorrect. Checking initial values is not a proof by induction.

(B) No, because PMI requires proving the inductive step ($P(k) \implies P(k+1)$). This is correct. Simply verifying the base case (and possibly other initial cases) is not enough; the crucial part of PMI is the general proof of the implication from $k$ to $k+1$ for all $k$ in the domain.

(C) Yes, because $n^2+n+41$ generates prime numbers for many initial values. This is incorrect. Generating primes for many values is interesting but does not prove it for *all* positive integers. As shown, it fails for $n=40$.

(D) No, because PMI can only be applied to formulas for sums or products. This is incorrect. PMI can be applied to various types of statements about integers, including inequalities and divisibility properties, not just sums or products.

The fact that $P(1)$ and $P(2)$ are true provides the base case for $n=1$ and some initial support, but it does not prove the statement for all positive integers. A complete PMI proof requires proving the inductive step.

The correct option is (B) No, because PMI requires proving the inductive step ($P(k) \implies P(k+1)$).

The Principle of Mathematical Induction (PMI) is logically equivalent to the Well-Ordering Principle for the set of positive integers (or natural numbers). The Well-Ordering Principle states that every non-empty subset of positive integers has a least element.

Let's see how PMI can be derived from the Well-Ordering Principle (or vice versa):

Assume PMI holds. Let $S$ be a non-empty subset of positive integers. We want to show $S$ has a least element. Assume for contradiction that $S$ has no least element. Let $P(n)$ be the statement "$n$ is not in $S$". $1$ must not be in $S$ (otherwise 1 would be the least element). So $P(1)$ is true. If $P(k)$ is true (i.e., $k \notin S$), then for $k+1$ to be in $S$, it would have to be the least element (since $1, 2, ..., k$ are not in $S$). But $S$ has no least element, so $k+1 \notin S$. Thus $P(k+1)$ is true. By PMI, $P(n)$ is true for all $n \geq 1$, meaning no positive integer is in $S$. This contradicts that $S$ is non-empty. So $S$ must have a least element.

Assume the Well-Ordering Principle holds. Let $P(n)$ be a statement such that $P(1)$ is true and $P(k) \implies P(k+1)$ for all $k \geq 1$. We want to show $P(n)$ is true for all $n \geq 1$. Assume for contradiction that there is some positive integer $n$ for which $P(n)$ is false. Let $S$ be the set of positive integers for which $P(n)$ is false. $S$ is non-empty by our assumption. By the Well-Ordering Principle, $S$ must have a least element, let's call it $m$. Since $P(1)$ is true, $1 \notin S$, so $m > 1$. Since $m$ is the least element in $S$, $m-1$ is a positive integer less than $m$, so $m-1 \notin S$. By definition of $S$, this means $P(m-1)$ is true. But the inductive step states that $P(m-1) \implies P(m)$ (since $m-1 \geq 1$ as $m > 1$). So, $P(m)$ must be true. This contradicts that $m \in S$ (meaning $P(m)$ is false). Therefore, the set $S$ must be empty, which means $P(n)$ is true for all positive integers $n$.

Let's look at the options:

(A) Real numbers being complete: This is a property of the set of real numbers (every Cauchy sequence converges, or every non-empty set bounded above has a least upper bound). This is not the basis for PMI.

(B) Every non-empty subset of positive integers having a least element: This is the Well-Ordering Principle, which is equivalent to PMI for the set of positive integers. This is the property on which PMI is based.

(C) Every sequence converging: This is related to analysis and limits, not the basis for standard PMI.

(D) The existence of an infinite number of primes: This is a theorem in number theory (Euclid's theorem), not the underlying principle for PMI.

The Principle of Mathematical Induction is a direct consequence of, or is equivalent to, the Well-Ordering Principle for the positive integers.

The correct option is (B) Every non-empty subset of positive integers having a least element.

The Principle of Mathematical Induction allows us to prove a statement $P(n)$ for all integers $n$ greater than or equal to the base case value $n_0$, provided two conditions are met:

1. Base Case: $P(n_0)$ is true.

2. Inductive Step: $P(k) \implies P(k+1)$ for all integers $k \geq n_0$.

In this problem, the statement is $P(n): 2^n > n^2$. We are told that the base case is proven for $n_0=5$, meaning $P(5)$ is true. We are also told that the inductive step $P(k) \implies P(k+1)$ is proven for all $k \geq 5$.

Applying PMI with base case $n_0=5$ and inductive step for $k \geq 5$:

• Since $P(5)$ is true (base case), and $P(5) \implies P(6)$ (inductive step for $k=5$), $P(6)$ is true.
• Since $P(6)$ is true, and $P(6) \implies P(7)$ (inductive step for $k=6$), $P(7)$ is true.
• This chain continues for all integers starting from 5.

Thus, the successful application of PMI with a base case at $n=5$ proves that the statement $P(n)$ is true for all integers $n$ such that $n \geq 5$.

Let's look at the options:

(A) For $n=1, 2, 3, 4$ only: The base case is at $n=5$, and the induction covers values $\geq 5$. This option is incorrect.

(B) For $n \geq 5$: This includes $n=5, 6, 7, ...$, which is the exact set of integers for which the statement is proven true by the described application of PMI.

(C) For all positive integers $n$: Positive integers are $1, 2, 3, 4, 5, 6, ...$. The induction starting at $n=5$ proves it for $n=5, 6, 7, ...$. It does not prove it for $n=1, 2, 3, 4$. (Indeed, as shown in a previous question, $P(2), P(3), P(4)$ are false). This option is incorrect.

(D) For no positive integers $n$: The base case $P(5)$ is stated as true, and the induction proves it for $n \geq 5$. This option is incorrect.

The specified PMI proof, with base case $n=5$ and inductive step for $k \geq 5$, proves the statement $P(n)$ for all integers greater than or equal to 5.

The correct option is (B) For $n \geq 5$.

The Principle of Mathematical Induction (PMI) is a proof technique for proving statements $P(n)$ about integers $n$ over a certain range, typically starting from a base case $n_0$. The key requirements are the Base Case and the Inductive Step.

Let's analyze the options in the context of proving $P(n)$ for all $n \geq n_0$:

(A) The statement must involve a formula for a sum: This is incorrect. PMI can be applied to various types of statements, including divisibility properties, inequalities, properties of sequences defined by recurrence relations, etc., not just summation formulas.

(B) The statement must be true for at least one positive integer: For PMI to prove the statement for $n \geq n_0$, the statement must be true for the specific base case $n_0$. If $n_0 \geq 1$, this means it must be true for at least one positive integer ($n_0$). However, this alone is not sufficient for the entire proof; the inductive step is also required. Also, if proving for $n \geq 0$, the base case is at 0, not necessarily a positive integer. While true for the base case is necessary, "at least one positive integer" might not be the full picture (if $n_0=0$) and is not a "key requirement" in isolation.

(C) The truth of $P(k)$ must logically imply the truth of $P(k+1)$ for $k$ in the relevant range: This is the core of the Inductive Step. For the principle to connect the base case to all subsequent integers, the implication $P(k) \implies P(k+1)$ must hold for all integers $k$ from the base case value $n_0$ onwards ($k \geq n_0$). This logical implication is a fundamental requirement of PMI.

(D) The statement must involve only integers: The statement $P(n)$ is a statement about the integer $n$. The statement itself can involve non-integers (e.g., $(1+x)^n \geq 1+nx$ involves a real number $x$), but the variable $n$ must be an integer that steps through the sequence. The properties or expressions within $P(n)$ don't strictly have to be composed *only* of integers, but the statement must be well-defined for integer values of $n$ in the relevant range.

Among the given options, the logical implication $P(k) \implies P(k+1)$ for $k$ in the relevant range (which constitutes the inductive step) is one of the two core pillars of a PMI proof (the other being the base case). Without a valid inductive step, the base case alone cannot prove the statement for all $n \geq n_0$. This implication is a key requirement for the principle to work.

The correct option is (C) The truth of $P(k)$ must logically imply the truth of $P(k+1)$ for $k$ in the relevant range.

The statement is $P(n): 1 + r + r^2 + ... + r^{n-1} = \frac{r^n - 1}{r - 1}$ (for $r \neq 1$), which is the formula for the sum of a geometric series with first term 1 and common ratio $r$. The sum on the LHS has $n$ terms.

The statement $P(k)$ is the formula for the sum of the first $k$ terms:

We need to prove $P(k+1)$ is true. The statement $P(k+1)$ is the formula for the sum of the first $k+1$ terms:

$P(k+1): 1 + r + ... + r^{k-1} + r^{(k+1)-1} = \frac{r^{k+1} - 1}{r - 1}$

$P(k+1): 1 + r + ... + r^{k-1} + r^k = \frac{r^{k+1} - 1}{r - 1}$

Let's analyze the options for what must be shown in the inductive step:

(A) $S_k + r^k = \frac{r^{k+1} - 1}{r - 1}$: Here $S_k$ represents the sum of the first $k$ terms, $1 + r + ... + r^{k-1}$. So, the LHS is the sum of the first $k$ terms plus the $(k+1)$-th term ($r^k$). The RHS is the formula for the sum of the first $k+1$ terms. This equality states that the sum of the first $k+1$ terms equals the expected formula for $n=k+1$. This is a correct representation of the goal of the inductive step.

(B) $\frac{r^k - 1}{r - 1} + r^k = \frac{r^{k+1} - 1}{r - 1}$: This is obtained from option (A) by substituting the inductive hypothesis $S_k = \frac{r^k - 1}{r - 1}$. This is the algebraic equation that needs to be verified to show that the inductive step holds. Showing that the left side equals the right side completes the proof of $P(k+1)$.

(C) $1 + r + ... + r^{k-1} + r^k = \frac{r^{k+1} - 1}{r - 1}$: This is the statement $P(k+1)$ itself. Proving $P(k+1)$ is true is the overall objective of the inductive step.

(D) All of the above are ways to represent the inductive step: Option (C) is the statement $P(k+1)$. Option (A) is the LHS of $P(k+1)$ written as the sum of the first $k$ terms plus the $(k+1)$-th term set equal to the RHS of $P(k+1)$. Option (B) is the algebraic equality derived from (A) by applying the inductive hypothesis. All three options describe aspects of what must be shown or manipulated in the inductive step. Proving (B) is the algebraic task that demonstrates (A) is true, which in turn demonstrates (C) is true. Therefore, all are valid representations related to showing $P(k+1)$ is true.

The correct option is (D) All of the above are ways to represent the inductive step.

The Principle of Mathematical Induction (PMI) is a proof technique specifically applicable to statements $P(n)$ where the variable $n$ belongs to the set of integers (or a subset of integers like natural numbers) and the property is proven for a sequence of these integers starting from a base case.

The described application of PMI (base case $P(1)$, inductive step $P(k) \implies P(k+1)$ for $k \geq 1$) proves that the statement $P(n)$ is true for all positive integers $n$, i.e., for $n \in \{1, 2, 3, ...\}$.

The question asks if this proof guarantees the truth of $P(1.5)$. The value $1.5$ is a real number that is not an integer.

PMI provides no information about the truth of the statement $P(n)$ for non-integer values of $n$, even if those non-integer values fall between integers for which the statement is proven. The method relies on the discrete, sequential structure of integers.

Unless the statement $P(n)$ is independently defined and proven for non-integer values of $n$, a proof by PMI for integer $n$ does not extend to non-integer values.

Let's look at the options:

(A) Yes, because $1.5$ is between 1 and 2, and the statement is proven for all integers $\geq 1$. This is incorrect. PMI does not fill in the gaps between integers.

(B) No, because PMI is a method for proving statements about integers, not real numbers. This is correct. The domain of application for standard PMI is integers.

(C) Yes, because the inductive step bridges the gap between $k$ and $k+1$. This is incorrect. The inductive step shows that if $P(k)$ is true, then $P(k+1)$ is true. It does not imply anything about values between $k$ and $k+1$.

(D) It depends on the specific statement $P(n)$. While for some specific statements $P(n)$, the corresponding statement for $n=1.5$ might happen to be true, the PMI proof for integer $n$ provides no logical guarantee for non-integer values. The question asks if the PMI proof *guarantees* the truth for $n=1.5$, and it does not. The truth for $n=1.5$ would require a separate proof or analysis.

The Principle of Mathematical Induction is a tool for proving statements about integers. A proof using PMI for $n \geq 1$ does not provide a guarantee for non-integer values like 1.5.

The correct option is (B) No, because PMI is a method for proving statements about integers, not real numbers.

The statement is $P(n): (ab)^n = a^n b^n$ for positive integers $n$ and real numbers $a, b$. We need to identify the base case $P(1)$.

The base case for proving this statement for all positive integers starting from $n=1$ is $P(1)$. To find $P(1)$, we substitute $n=1$ into the statement $P(n)$.

$P(1): (ab)^1 = a^1 b^1$

Let's evaluate both sides of the equality for $n=1$:

Left side: $(ab)^1 = ab$

Right side: $a^1 b^1 = a b$

So, $P(1)$ is the statement $ab = ab$. This is a true statement.

Let's look at the options:

(A) $(ab)^1 = a^1 b^1$, which is $ab = ab$. True: This option states the equality for $n=1$ and shows the simplification, concluding that the statement $P(1)$ is true. This is a correct and complete description of the base case $P(1)$ verification.

(B) $(a \cdot b)^1 = a^1 \cdot b^1$: This is the statement $P(1)$ itself, written with the dot notation for multiplication.

(C) The statement is true for $n=1$: This states the truth value of $P(1)$ but doesn't explicitly show the statement $P(1)$.

(D) All of the above describe the base case $P(1)$: Option (B) is the statement $P(1)$. Option (A) shows the statement $P(1)$, simplifies it, and verifies its truth. Option (C) states that $P(1)$ is true. All three options relate to the base case $P(1)$. Option (B) gives the form of $P(1)$. Option (A) gives the form, simplifies, and verifies its truth. Option (C) gives the truth value. All are aspects of considering the base case $P(1)$.

Option (B) is the statement $P(1)$. Option (A) is the verification of $P(1)$, which includes stating $P(1)$ and checking its truth. Option (C) is the conclusion of the verification. All are related to the base case $P(1)$ and its truth. Given the options, (D) is the most appropriate answer, as (A), (B), and (C) collectively describe the base case and its verification.

The correct option is (D) All of the above describe the base case $P(1)$.

Given:

A statement $P(n)$ concerning the sum of the first $n$ odd natural numbers.

$P(n): 1 + 3 + 5 + \dots + (2n-1) = n^2$, for all $n \in \mathbb{N}$.

To Prove:

The statement $P(n)$ is true for all natural numbers $n$, using the Principle of Mathematical Induction.

Proof:

We prove the statement $P(n)$ using the Principle of Mathematical Induction. The principle involves two steps:

Step 1: Base Case

We verify if the statement is true for $n=1$.

For $n=1$, the left-hand side (LHS) of $P(n)$ is the sum of the first odd number, which is $1$.

$LHS = 1$

The right-hand side (RHS) of $P(n)$ for $n=1$ is $n^2$.

$RHS = 1^2 = 1$

Since $LHS = RHS$, the statement $P(1)$ is true.

Step 2: Inductive Hypothesis

Assume that the statement $P(k)$ is true for some arbitrary positive integer $k$. That is, we assume:

$1 + 3 + 5 + \dots + (2k-1) = k^2$

Step 3: Inductive Step

We need to prove that the statement $P(k+1)$ is true whenever $P(k)$ is true. That is, we need to prove that:

$1 + 3 + 5 + \dots + (2k-1) + (2(k+1)-1) = (k+1)^2$

Consider the LHS of $P(k+1)$:

$LHS = 1 + 3 + 5 + \dots + (2k-1) + (2(k+1)-1)$

We can group the first $k$ terms, which is the LHS of $P(k)$.

$LHS = [1 + 3 + 5 + \dots + (2k-1)] + (2(k+1)-1)$

Using the Inductive Hypothesis from equation (i), we substitute $k^2$ for the sum of the first $k$ terms:

$LHS = k^2 + (2(k+1)-1)$

Now, simplify the term $(2(k+1)-1)$:

$2(k+1)-1 = 2k + 2 - 1 = 2k + 1$

So, the LHS becomes:

$LHS = k^2 + 2k + 1$

This expression is a perfect square trinomial, which can be factored as $(k+1)^2$.

$LHS = (k+1)^2$

This is the RHS of the statement $P(k+1)$.

Thus, we have shown that $LHS = RHS$, which means $P(k+1)$ is true whenever $P(k)$ is true.

Conclusion:

Since $P(1)$ is true (Base Case) and $P(k+1)$ is true whenever $P(k)$ is true (Inductive Step), by the Principle of Mathematical Induction, the statement $P(n): 1 + 3 + 5 + \dots + (2n-1) = n^2$ is true for all natural numbers $n \in \mathbb{N}$.

Given:

A statement $P(n)$ concerning the sum of the squares of the first $n$ natural numbers.

$P(n): 1^2 + 2^2 + \dots + n^2 = \frac{n(n+1)(2n+1)}{6}$, for all $n \in \mathbb{N}$.

To Prove:

The statement $P(n)$ is true for all natural numbers $n$, using the Principle of Mathematical Induction.

Proof:

We prove the statement $P(n)$ using the Principle of Mathematical Induction. The principle involves two steps:

Step 1: Base Case

We verify if the statement is true for $n=1$.

For $n=1$, the left-hand side (LHS) of $P(n)$ is the square of the first natural number, which is $1^2$.

$LHS = 1^2 = 1$

The right-hand side (RHS) of $P(n)$ for $n=1$ is $\frac{1(1+1)(2(1)+1)}{6}$.

$RHS = \frac{1(2)(3)}{6} = \frac{6}{6} = 1$

Since $LHS = RHS$, the statement $P(1)$ is true.

Step 2: Inductive Hypothesis

Assume that the statement $P(k)$ is true for some arbitrary positive integer $k$. That is, we assume:

$1^2 + 2^2 + \dots + k^2 = \frac{k(k+1)(2k+1)}{6}$

Step 3: Inductive Step

We need to prove that the statement $P(k+1)$ is true whenever $P(k)$ is true. That is, we need to prove that:

$$1^2 + 2^2 + \dots + k^2 + (k+1)^2 = \frac{(k+1)((k+1)+1)(2(k+1)+1)}{6}$$

$$1^2 + 2^2 + \dots + k^2 + (k+1)^2 = \frac{(k+1)(k+2)(2k+3)}{6}$$

Consider the LHS of $P(k+1)$:

$$LHS = 1^2 + 2^2 + \dots + k^2 + (k+1)^2$$

We can group the first $k$ terms, which is the LHS of $P(k)$.

$$LHS = [1^2 + 2^2 + \dots + k^2] + (k+1)^2$$

Using the Inductive Hypothesis from equation (i), we substitute $\frac{k(k+1)(2k+1)}{6}$ for the sum of the first $k$ terms:

$$LHS = \frac{k(k+1)(2k+1)}{6} + (k+1)^2$$

To combine the terms, find a common denominator:

$$LHS = \frac{k(k+1)(2k+1)}{6} + \frac{6(k+1)^2}{6}$$

Combine the numerators:

$$LHS = \frac{k(k+1)(2k+1) + 6(k+1)^2}{6}$$

Factor out the common term $(k+1)$ from the numerator:

$$LHS = \frac{(k+1) [k(2k+1) + 6(k+1)]}{6}$$

Simplify the expression inside the square brackets:

$$k(2k+1) + 6(k+1) = 2k^2 + k + 6k + 6 = 2k^2 + 7k + 6$$

Now, substitute this back into the expression for LHS:

$$LHS = \frac{(k+1) (2k^2 + 7k + 6)}{6}$$

Factor the quadratic $2k^2 + 7k + 6$. We look for two numbers that multiply to $2 \times 6 = 12$ and add up to $7$. These numbers are $3$ and $4$.

$$2k^2 + 7k + 6 = 2k^2 + 3k + 4k + 6$$

$$= k(2k+3) + 2(2k+3)$$

$$= (k+2)(2k+3)$$

Substitute the factored quadratic back into the LHS expression:

$$LHS = \frac{(k+1) (k+2)(2k+3)}{6}$$

This is the RHS of the statement $P(k+1)$.

Thus, we have shown that $LHS = RHS$, which means $P(k+1)$ is true whenever $P(k)$ is true.

Conclusion:

Since $P(1)$ is true (Base Case) and $P(k+1)$ is true whenever $P(k)$ is true (Inductive Step), by the Principle of Mathematical Induction, the statement $P(n): 1^2 + 2^2 + \dots + n^2 = \frac{n(n+1)(2n+1)}{6}$ is true for all natural numbers $n \in \mathbb{N}$.

Given:

A statement $P(n)$ concerning an inequality involving $n$ and $2^n$.

$P(n): 2^n > n$, for all positive integers $n \in \mathbb{N}$.

To Prove:

The statement $P(n)$ is true for all positive integers $n$, using the Principle of Mathematical Induction.

Proof:

We prove the statement $P(n)$ using the Principle of Mathematical Induction. The principle involves two steps:

Step 1: Base Case

We verify if the statement is true for the smallest positive integer, $n=1$.

For $n=1$, the LHS of $P(n)$ is $2^1$.

$LHS = 2^1 = 2$

The RHS of $P(n)$ for $n=1$ is $n$.

$RHS = 1$

Since $2 > 1$, which is $LHS > RHS$, the statement $P(1)$ is true.

Step 2: Inductive Hypothesis

Assume that the statement $P(k)$ is true for some arbitrary positive integer $k \geq 1$. That is, we assume:

Step 3: Inductive Step

We need to prove that the statement $P(k+1)$ is true whenever $P(k)$ is true. That is, we need to prove that:

$$2^{k+1} > k+1$$

Consider the LHS of $P(k+1)$:

$$LHS = 2^{k+1}$$

We can write $2^{k+1}$ as $2 \cdot 2^k$.

$$LHS = 2 \cdot 2^k$$

Using the Inductive Hypothesis from inequality (i), we know that $2^k > k$. Multiplying both sides of this inequality by $2$ (which is a positive number, so the inequality direction is preserved):

So,

Now, we need to show that $2k \geq k+1$ for $k \geq 1$.

Consider the difference $2k - (k+1)$:

$$2k - (k+1) = 2k - k - 1 = k - 1$$

Since $k$ is a positive integer and $k \geq 1$, we have $k-1 \geq 0$.

This implies:

Therefore,

Combining the inequalities $2^{k+1} > 2k$ and $2k \geq k+1$, we can conclude that:

$$2^{k+1} > 2k \geq k+1$$

By the transitive property of inequality, this gives us:

$$2^{k+1} > k+1$$

This is the RHS of the statement $P(k+1)$.

Thus, we have shown that $P(k+1)$ is true whenever $P(k)$ is true.

Conclusion:

Since $P(1)$ is true (Base Case) and $P(k+1)$ is true whenever $P(k)$ is true for $k \geq 1$ (Inductive Step), by the Principle of Mathematical Induction, the statement $P(n): 2^n > n$ is true for all positive integers $n \in \mathbb{N}$.

Given:

A statement $P(n)$ concerning the divisibility of an expression by 64.

$P(n): 3^{2n+2} - 8n - 9$ is divisible by 64, for all integers $n \geq 1$.

To Prove:

The statement $P(n)$ is true for all integers $n \geq 1$, using the Principle of Mathematical Induction.

Proof:

We prove the statement $P(n)$ using the Principle of Mathematical Induction. The principle involves two steps:

Step 1: Base Case

We verify if the statement is true for the smallest value of $n$, which is $n=1$.

For $n=1$, we evaluate the expression $3^{2(1)+2} - 8(1) - 9$:

$$3^{2(1)+2} - 8(1) - 9 = 3^{2+2} - 8 - 9$$

$$= 3^4 - 17$$

$$= 81 - 17$$

$$= 64$$

Since $64$ is divisible by $64$, the statement $P(1)$ is true.

Step 2: Inductive Hypothesis

Assume that the statement $P(k)$ is true for some arbitrary positive integer $k \geq 1$.

This means that $3^{2k+2} - 8k - 9$ is divisible by 64. Therefore, we can write:

where $m$ is some integer.

Step 3: Inductive Step

We need to prove that the statement $P(k+1)$ is true whenever $P(k)$ is true. That is, we need to prove that $3^{2(k+1)+2} - 8(k+1) - 9$ is divisible by 64.

Consider the expression for $P(k+1)$:

$$3^{2(k+1)+2} - 8(k+1) - 9$$

$$= 3^{2k+2+2} - 8k - 8 - 9$$

$$= 3^{2k+4} - 8k - 17$$

We can rewrite $3^{2k+4}$ as $3^2 \cdot 3^{2k+2}$:

$$= 9 \cdot 3^{2k+2} - 8k - 17$$

From the Inductive Hypothesis (equation i), we know that $3^{2k+2} = 64m + 8k + 9$. Substitute this into the expression:

$$= 9 \cdot (64m + 8k + 9) - 8k - 17$$

Distribute the 9:

$$= 9 \cdot 64m + 9 \cdot 8k + 9 \cdot 9 - 8k - 17$$

$$= 9 \cdot 64m + 72k + 81 - 8k - 17$$

Combine the terms with $k$ and the constant terms:

$$= 9 \cdot 64m + (72k - 8k) + (81 - 17)$$

$$= 9 \cdot 64m + 64k + 64$$

Now, factor out 64 from the expression:

$$= 64 (9m + k + 1)$$

Since $m$ is an integer (from the Inductive Hypothesis) and $k$ is an integer, the term $(9m + k + 1)$ is also an integer.

Therefore, $3^{2(k+1)+2} - 8(k+1) - 9$ is a multiple of 64, which means it is divisible by 64.

Thus, we have shown that $P(k+1)$ is true whenever $P(k)$ is true.

Conclusion:

Since $P(1)$ is true (Base Case) and $P(k+1)$ is true whenever $P(k)$ is true for $k \geq 1$ (Inductive Step), by the Principle of Mathematical Induction, the statement $P(n): 3^{2n+2} - 8n - 9$ is divisible by 64 for all integers $n \geq 1$.

Given:

A statement $P(n)$ concerning the sum of the product of consecutive natural numbers.

$$P(n): \sum_{i=1}^{n} i(i+1) = 1 \cdot 2 + 2 \cdot 3 + \dots + n(n+1) = \frac{n(n+1)(n+2)}{3}$$

for all $n \in \mathbb{N}$.

To Prove:

The statement $P(n)$ is true for all natural numbers $n$, using the Principle of Mathematical Induction.

Proof:

We prove the statement $P(n)$ using the Principle of Mathematical Induction. The principle involves two steps:

Step 1: Base Case

We verify if the statement is true for $n=1$.

For $n=1$, the left-hand side (LHS) of $P(n)$ is the first term of the sum.

$$LHS = 1 \cdot (1+1) = 1 \cdot 2 = 2$$

The right-hand side (RHS) of $P(n)$ for $n=1$ is $\frac{1(1+1)(1+2)}{3}$.

$$RHS = \frac{1(2)(3)}{3} = \frac{6}{3} = 2$$

Since $LHS = RHS$, the statement $P(1)$ is true.

Step 2: Inductive Hypothesis

Assume that the statement $P(k)$ is true for some arbitrary positive integer $k$. That is, we assume:

Step 3: Inductive Step

We need to prove that the statement $P(k+1)$ is true whenever $P(k)$ is true. That is, we need to prove that:

$$1 \cdot 2 + 2 \cdot 3 + \dots + k(k+1) + (k+1)((k+1)+1) = \frac{(k+1)((k+1)+1)((k+1)+2)}{3}$$

$$1 \cdot 2 + 2 \cdot 3 + \dots + k(k+1) + (k+1)(k+2) = \frac{(k+1)(k+2)(k+3)}{3}$$

Consider the LHS of $P(k+1)$:

$$LHS = 1 \cdot 2 + 2 \cdot 3 + \dots + k(k+1) + (k+1)(k+2)$$

We can group the first $k$ terms, which is the LHS of $P(k)$.

$$LHS = [1 \cdot 2 + 2 \cdot 3 + \dots + k(k+1)] + (k+1)(k+2)$$

Using the Inductive Hypothesis from equation (i), we substitute $\frac{k(k+1)(k+2)}{3}$ for the sum of the first $k$ terms:

$$LHS = \frac{k(k+1)(k+2)}{3} + (k+1)(k+2)$$

To combine the terms, find a common denominator:

$$LHS = \frac{k(k+1)(k+2)}{3} + \frac{3(k+1)(k+2)}{3}$$

Combine the numerators:

$$LHS = \frac{k(k+1)(k+2) + 3(k+1)(k+2)}{3}$$

Factor out the common term $(k+1)(k+2)$ from the numerator:

$$LHS = \frac{(k+1)(k+2) [k + 3]}{3}$$

Rearrange the terms in the numerator:

$$LHS = \frac{(k+1)(k+2)(k+3)}{3}$$

This is the RHS of the statement $P(k+1)$.

Thus, we have shown that $LHS = RHS$, which means $P(k+1)$ is true whenever $P(k)$ is true.

Conclusion:

Since $P(1)$ is true (Base Case) and $P(k+1)$ is true whenever $P(k)$ is true (Inductive Step), by the Principle of Mathematical Induction, the statement $P(n): 1 \cdot 2 + 2 \cdot 3 + \dots + n(n+1) = \frac{n(n+1)(n+2)}{3}$ is true for all natural numbers $n \in \mathbb{N}$.

Given:

A statement $P(n)$ concerning an inequality.

$P(n): 4^n < 2^{2n+1}$, for all $n \in \mathbb{N}$.

To Prove:

The statement $P(n)$ is true for all natural numbers $n$, using the Principle of Mathematical Induction.

Proof:

We prove the statement $P(n)$ using the Principle of Mathematical Induction. The principle involves two steps:

Step 1: Base Case

We verify if the statement is true for the smallest natural number, $n=1$.

For $n=1$, the left-hand side (LHS) of $P(n)$ is $4^1$.

$$LHS = 4^1 = 4$$

The right-hand side (RHS) of $P(n)$ for $n=1$ is $2^{2(1)+1}$.

$$RHS = 2^{2+1} = 2^3 = 8$$

Since $4 < 8$, which is $LHS < RHS$, the statement $P(1)$ is true.

Step 2: Inductive Hypothesis

Assume that the statement $P(k)$ is true for some arbitrary positive integer $k \geq 1$. That is, we assume:

Step 3: Inductive Step

We need to prove that the statement $P(k+1)$ is true whenever $P(k)$ is true. That is, we need to prove that:

$$4^{k+1} < 2^{2(k+1)+1}$$

$$4^{k+1} < 2^{2k+3}$$

Consider the LHS of $P(k+1)$:

$$LHS = 4^{k+1}$$

We can write $4^{k+1}$ as $4^k \cdot 4$.

$$LHS = 4 \cdot 4^k$$

Using the Inductive Hypothesis from inequality (i), we know that $4^k < 2^{2k+1}$. Multiply both sides of this inequality by $4$ (which is a positive number, so the inequality direction is preserved):

$$4^{k+1} < 4 \cdot 2^{2k+1}$$

We can express $4$ as $2^2$. Substitute this into the inequality:

$$4^{k+1} < 2^2 \cdot 2^{2k+1}$$

Using the property of exponents $a^m \cdot a^n = a^{m+n}$, we combine the terms on the RHS:

$$4^{k+1} < 2^{2 + (2k+1)}$$

$$4^{k+1} < 2^{2k+3}$$

This is the RHS of the statement $P(k+1)$.

Thus, we have shown that $LHS < RHS$, which means $P(k+1)$ is true whenever $P(k)$ is true.

Conclusion:

Since $P(1)$ is true (Base Case) and $P(k+1)$ is true whenever $P(k)$ is true for $k \geq 1$ (Inductive Step), by the Principle of Mathematical Induction, the statement $P(n): 4^n < 2^{2n+1}$ is true for all natural numbers $n \in \mathbb{N}$.

Given:

A statement $P(n)$ concerning the divisibility of an expression by 4.

$P(n): 7^n - 3^n$ is divisible by 4, for all natural numbers $n \in \mathbb{N}$.

To Prove:

The statement $P(n)$ is true for all natural numbers $n$, using the Principle of Mathematical Induction.

Proof:

We prove the statement $P(n)$ using the Principle of Mathematical Induction. The principle involves two steps:

Step 1: Base Case

We verify if the statement is true for the smallest natural number, $n=1$.

For $n=1$, we evaluate the expression $7^1 - 3^1$:

$$7^1 - 3^1 = 7 - 3 = 4$$

Since $4$ is divisible by $4$, the statement $P(1)$ is true.

Step 2: Inductive Hypothesis

Assume that the statement $P(k)$ is true for some arbitrary positive integer $k \geq 1$.

This means that $7^k - 3^k$ is divisible by 4. Therefore, we can write:

where $m$ is some integer.

From equation (i), we can express $7^k$ as:

Step 3: Inductive Step

We need to prove that the statement $P(k+1)$ is true whenever $P(k)$ is true. That is, we need to prove that $7^{k+1} - 3^{k+1}$ is divisible by 4.

Consider the expression for $P(k+1)$:

$$7^{k+1} - 3^{k+1}$$

We can rewrite this as $7 \cdot 7^k - 3 \cdot 3^k$. Substitute the expression for $7^k$ from the Inductive Hypothesis ($7^k = 4m + 3^k$):

$$7^{k+1} - 3^{k+1} = 7 \cdot (4m + 3^k) - 3 \cdot 3^k$$

Distribute the 7:

$$= 7 \cdot 4m + 7 \cdot 3^k - 3 \cdot 3^k$$

Group the terms involving $3^k$:

$$= 28m + (7 - 3) \cdot 3^k$$

$$= 28m + 4 \cdot 3^k$$

Now, factor out 4 from the expression:

$$= 4 (7m + 3^k)$$

Since $m$ is an integer (from the Inductive Hypothesis) and $k$ is a positive integer, $3^k$ is an integer. Therefore, the term $(7m + 3^k)$ is also an integer.

Let $p = 7m + 3^k$, where $p$ is an integer. Then we have:

$$7^{k+1} - 3^{k+1} = 4p$$

This shows that $7^{k+1} - 3^{k+1}$ is a multiple of 4, which means it is divisible by 4.

Thus, we have shown that $P(k+1)$ is true whenever $P(k)$ is true.

Conclusion:

Since $P(1)$ is true (Base Case) and $P(k+1)$ is true whenever $P(k)$ is true for $k \geq 1$ (Inductive Step), by the Principle of Mathematical Induction, the statement $P(n): 7^n - 3^n$ is divisible by 4 is true for all natural numbers $n \in \mathbb{N}$.

Given:

A statement $P(n)$ concerning the sum of the product of a natural number and that number plus two.

$$P(n): 1 \cdot 3 + 2 \cdot 4 + \dots + n(n+2) = \frac{n(n+1)(2n+7)}{6}$$

for all $n \in \mathbb{N}$.

To Prove:

The statement $P(n)$ is true for all natural numbers $n$, using the Principle of Mathematical Induction.

Proof:

We prove the statement $P(n)$ using the Principle of Mathematical Induction. The principle involves two steps:

Step 1: Base Case

We verify if the statement is true for $n=1$.

For $n=1$, the left-hand side (LHS) of $P(n)$ is the first term of the sum.

$$LHS = 1 \cdot (1+2) = 1 \cdot 3 = 3$$

The right-hand side (RHS) of $P(n)$ for $n=1$ is $\frac{1(1+1)(2(1)+7)}{6}$.

$$RHS = \frac{1(2)(2+7)}{6} = \frac{1(2)(9)}{6} = \frac{18}{6} = 3$$

Since $LHS = RHS$, the statement $P(1)$ is true.

Step 2: Inductive Hypothesis

Assume that the statement $P(k)$ is true for some arbitrary positive integer $k$. That is, we assume:

Step 3: Inductive Step

We need to prove that the statement $P(k+1)$ is true whenever $P(k)$ is true. That is, we need to prove that:

$$1 \cdot 3 + 2 \cdot 4 + \dots + k(k+2) + (k+1)((k+1)+2) = \frac{(k+1)((k+1)+1)(2(k+1)+7)}{6}$$

$$1 \cdot 3 + 2 \cdot 4 + \dots + k(k+2) + (k+1)(k+3) = \frac{(k+1)(k+2)(2k+2+7)}{6}$$

$$1 \cdot 3 + 2 \cdot 4 + \dots + k(k+2) + (k+1)(k+3) = \frac{(k+1)(k+2)(2k+9)}{6}$$

Consider the LHS of $P(k+1)$:

$$LHS = 1 \cdot 3 + 2 \cdot 4 + \dots + k(k+2) + (k+1)(k+3)$$

We can group the first $k$ terms, which is the LHS of $P(k)$.

$$LHS = [1 \cdot 3 + 2 \cdot 4 + \dots + k(k+2)] + (k+1)(k+3)$$

Using the Inductive Hypothesis from equation (i), we substitute $\frac{k(k+1)(2k+7)}{6}$ for the sum of the first $k$ terms:

$$LHS = \frac{k(k+1)(2k+7)}{6} + (k+1)(k+3)$$

To combine the terms, find a common denominator:

$$LHS = \frac{k(k+1)(2k+7)}{6} + \frac{6(k+1)(k+3)}{6}$$

Combine the numerators:

$$LHS = \frac{k(k+1)(2k+7) + 6(k+1)(k+3)}{6}$$

Factor out the common term $(k+1)$ from the numerator:

$$LHS = \frac{(k+1) [k(2k+7) + 6(k+3)]}{6}$$

Simplify the expression inside the square brackets:

$$k(2k+7) + 6(k+3) = 2k^2 + 7k + 6k + 18 = 2k^2 + 13k + 18$$

Now, substitute this back into the expression for LHS:

$$LHS = \frac{(k+1) (2k^2 + 13k + 18)}{6}$$

Factor the quadratic $2k^2 + 13k + 18$. We look for two numbers that multiply to $2 \times 18 = 36$ and add up to $13$. These numbers are $4$ and $9$.

$$2k^2 + 13k + 18 = 2k^2 + 4k + 9k + 18$$

$$= 2k(k+2) + 9(k+2)$$

$$= (2k+9)(k+2)$$

Substitute the factored quadratic back into the LHS expression:

$$LHS = \frac{(k+1) (k+2)(2k+9)}{6}$$

This is the RHS of the statement $P(k+1)$.

Thus, we have shown that $LHS = RHS$, which means $P(k+1)$ is true whenever $P(k)$ is true.

Conclusion:

Since $P(1)$ is true (Base Case) and $P(k+1)$ is true whenever $P(k)$ is true (Inductive Step), by the Principle of Mathematical Induction, the statement $P(n): 1 \cdot 3 + 2 \cdot 4 + \dots + n(n+2) = \frac{n(n+1)(2n+7)}{6}$ is true for all natural numbers $n \in \mathbb{N}$.

Given:

A statement $P(n)$ concerning the sum of the cubes of the first $n$ natural numbers.

$$P(n): \sum\limits_{i=1}^{n} i^3 = 1^3 + 2^3 + \dots + n^3 = \left(\frac{n(n+1)}{2}\right)^2$$

for all $n \in \mathbb{N}$.

To Prove:

The statement $P(n)$ is true for all natural numbers $n$, using the Principle of Mathematical Induction.

Proof:

We prove the statement $P(n)$ using the Principle of Mathematical Induction. The principle involves two steps:

Step 1: Base Case

We verify if the statement is true for $n=1$.

For $n=1$, the left-hand side (LHS) of $P(n)$ is the first term of the sum.

$$LHS = 1^3 = 1$$

The right-hand side (RHS) of $P(n)$ for $n=1$ is $\left(\frac{1(1+1)}{2}\right)^2$.

$$RHS = \left(\frac{1(2)}{2}\right)^2 = \left(\frac{2}{2}\right)^2 = 1^2 = 1$$

Since $LHS = RHS$, the statement $P(1)$ is true.

Step 2: Inductive Hypothesis

Assume that the statement $P(k)$ is true for some arbitrary positive integer $k$. That is, we assume:

Step 3: Inductive Step

We need to prove that the statement $P(k+1)$ is true whenever $P(k)$ is true. That is, we need to prove that:

$$1^3 + 2^3 + \dots + k^3 + (k+1)^3 = \left(\frac{(k+1)((k+1)+1)}{2}\right)^2$$

$$1^3 + 2^3 + \dots + k^3 + (k+1)^3 = \left(\frac{(k+1)(k+2)}{2}\right)^2$$

Consider the LHS of $P(k+1)$:

$$LHS = 1^3 + 2^3 + \dots + k^3 + (k+1)^3$$

We can group the first $k$ terms, which is the LHS of $P(k)$.

$$LHS = [1^3 + 2^3 + \dots + k^3] + (k+1)^3$$

Using the Inductive Hypothesis from equation (i), we substitute $\left(\frac{k(k+1)}{2}\right)^2$ for the sum of the first $k$ terms:

$$LHS = \left(\frac{k(k+1)}{2}\right)^2 + (k+1)^3$$

$$LHS = \frac{k^2(k+1)^2}{4} + (k+1)^3$$

To combine the terms, find a common denominator:

$$LHS = \frac{k^2(k+1)^2}{4} + \frac{4(k+1)^3}{4}$$

Combine the numerators:

$$LHS = \frac{k^2(k+1)^2 + 4(k+1)^3}{4}$$

Factor out the common term $(k+1)^2$ from the numerator:

$$LHS = \frac{(k+1)^2 [k^2 + 4(k+1)]}{4}$$

Simplify the expression inside the square brackets:

$$k^2 + 4(k+1) = k^2 + 4k + 4$$

This is a perfect square trinomial, which can be factored as $(k+2)^2$.

$$k^2 + 4k + 4 = (k+2)^2$$

Now, substitute this back into the expression for LHS:

$$LHS = \frac{(k+1)^2 (k+2)^2}{4}$$

This expression can be written as:

$$LHS = \left(\frac{(k+1)(k+2)}{2}\right)^2$$

This is the RHS of the statement $P(k+1)$.

Thus, we have shown that $LHS = RHS$, which means $P(k+1)$ is true whenever $P(k)$ is true.

Conclusion:

Since $P(1)$ is true (Base Case) and $P(k+1)$ is true whenever $P(k)$ is true (Inductive Step), by the Principle of Mathematical Induction, the statement $P(n): \sum\limits_{i=1}^{n} i^3 = \left(\frac{n(n+1)}{2}\right)^2$ is true for all natural numbers $n \in \mathbb{N}$.

Given:

A statement $P(n)$ concerning the divisibility of an expression by 6.

$P(n): n^3 + n$ is divisible by 6, for all natural numbers $n \in \mathbb{N}$.

To Prove:

The statement $P(n)$ is true for all natural numbers $n$, using the Principle of Mathematical Induction.

Proof:

We attempt to prove the statement $P(n)$ using the Principle of Mathematical Induction. The principle involves two steps:

Step 1: Base Case

We verify if the statement is true for the smallest natural number, $n=1$.

For $n=1$, we evaluate the expression $1^3 + 1$:

$$1^3 + 1 = 1 + 1 = 2$$

We check if this value is divisible by 6.

$$2 \div 6 = \frac{2}{6} = \frac{1}{3}$$

Since the result is not an integer, 2 is not divisible by 6.

Thus, the statement $P(1)$ is false.

Conclusion:

The statement $P(n): n^3 + n$ is divisible by 6 for all $n \in \mathbb{N}$ is false, as shown by the failure of the Base Case ($P(1)$ is false). The Principle of Mathematical Induction can only be used to prove statements that are true for the specified range of numbers. Since the statement is false for $n=1$, it is not true for all $n \in \mathbb{N}$, and therefore, cannot be proven true by the Principle of Mathematical Induction.

Given:

A statement $P(n)$ concerning the sum of a series of fractions.

$$P(n): \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{n(n+1)} = \frac{n}{n+1}$$

for all $n \in \mathbb{N}$.

To Prove:

The statement $P(n)$ is true for all natural numbers $n$, using the Principle of Mathematical Induction.

Proof:

We prove the statement $P(n)$ using the Principle of Mathematical Induction. The principle involves two steps:

Step 1: Base Case

We verify if the statement is true for the smallest natural number, $n=1$.

For $n=1$, the left-hand side (LHS) of $P(n)$ is the first term of the sum.

$$LHS = \frac{1}{1 \cdot (1+1)} = \frac{1}{1 \cdot 2} = \frac{1}{2}$$

The right-hand side (RHS) of $P(n)$ for $n=1$ is $\frac{1}{1+1}$.

$$RHS = \frac{1}{1+1} = \frac{1}{2}$$

Since $LHS = RHS$, the statement $P(1)$ is true.

Step 2: Inductive Hypothesis

Assume that the statement $P(k)$ is true for some arbitrary positive integer $k$. That is, we assume:

Step 3: Inductive Step

We need to prove that the statement $P(k+1)$ is true whenever $P(k)$ is true. That is, we need to prove that:

$$\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{k(k+1)} + \frac{1}{(k+1)((k+1)+1)} = \frac{k+1}{(k+1)+1}$$

$$\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{k(k+1)} + \frac{1}{(k+1)(k+2)} = \frac{k+1}{k+2}$$

Consider the LHS of $P(k+1)$:

$$LHS = \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{k(k+1)} + \frac{1}{(k+1)(k+2)}$$

We can group the first $k$ terms, which is the LHS of $P(k)$.

$$LHS = \left[\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{k(k+1)}\right] + \frac{1}{(k+1)(k+2)}$$

Using the Inductive Hypothesis from equation (i), we substitute $\frac{k}{k+1}$ for the sum of the first $k$ terms:

$$LHS = \frac{k}{k+1} + \frac{1}{(k+1)(k+2)}$$

To combine the fractions, find a common denominator, which is $(k+1)(k+2)$:

$$LHS = \frac{k(k+2)}{(k+1)(k+2)} + \frac{1}{(k+1)(k+2)}$$

Combine the numerators:

$$LHS = \frac{k(k+2) + 1}{(k+1)(k+2)}$$

Simplify the numerator:

$$k(k+2) + 1 = k^2 + 2k + 1$$

This is a perfect square trinomial, which can be factored as $(k+1)^2$:

$$k^2 + 2k + 1 = (k+1)^2$$

Substitute the factored numerator back into the LHS expression:

$$LHS = \frac{(k+1)^2}{(k+1)(k+2)}$$

Assuming $k+1 \neq 0$ (which is true since $k \in \mathbb{N}$), we can cancel one factor of $(k+1)$ from the numerator and denominator:

$$LHS = \frac{\cancel{(k+1)}^{\,1} (k+1)}{\cancel{(k+1)}_{\,1} (k+2)} = \frac{k+1}{k+2}$$

This is the RHS of the statement $P(k+1)$.

Thus, we have shown that $LHS = RHS$, which means $P(k+1)$ is true whenever $P(k)$ is true.

Conclusion:

Since $P(1)$ is true (Base Case) and $P(k+1)$ is true whenever $P(k)$ is true (Inductive Step), by the Principle of Mathematical Induction, the statement $P(n): \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{n(n+1)} = \frac{n}{n+1}$ is true for all natural numbers $n \in \mathbb{N}$.

Given:

A statement $P(n)$ concerning the divisibility of an expression by 3.

$P(n): n(n+1)(n+5)$ is divisible by 3, for all natural numbers $n \in \mathbb{N}$.

To Prove:

The statement $P(n)$ is true for all natural numbers $n$, using the Principle of Mathematical Induction.

Proof:

We prove the statement $P(n)$ using the Principle of Mathematical Induction. The principle involves two steps:

Step 1: Base Case

We verify if the statement is true for the smallest natural number, $n=1$.

For $n=1$, we evaluate the expression $1(1+1)(1+5)$:

$$1(1+1)(1+5) = 1(2)(6) = 12$$

We check if $12$ is divisible by 3.

$$12 \div 3 = 4$$

Since 12 is divisible by 3, the statement $P(1)$ is true.

Step 2: Inductive Hypothesis

Assume that the statement $P(k)$ is true for some arbitrary positive integer $k \geq 1$.

This means that $k(k+1)(k+5)$ is divisible by 3. Therefore, we can write:

where $m$ is some integer.

Step 3: Inductive Step

We need to prove that the statement $P(k+1)$ is true whenever $P(k)$ is true. That is, we need to prove that $(k+1)((k+1)+1)((k+1)+5)$ is divisible by 3.

Consider the expression for $P(k+1)$:

$$(k+1)((k+1)+1)((k+1)+5) = (k+1)(k+2)(k+6)$$

Let's expand this expression:

$$(k+1)(k+2)(k+6) = (k^2 + 2k + k + 2)(k+6)$$

$$= (k^2 + 3k + 2)(k+6)$$

$$= k^2(k+6) + 3k(k+6) + 2(k+6)$$

$$= k^3 + 6k^2 + 3k^2 + 18k + 2k + 12$$

$$= k^3 + 9k^2 + 20k + 12$$

From the Inductive Hypothesis (equation i), we have $k(k+1)(k+5) = k(k^2 + 6k + 5) = k^3 + 6k^2 + 5k = 3m$.

We can rewrite $k^3$ from the inductive hypothesis as $k^3 = 3m - 6k^2 - 5k$. Substitute this into the expression for $P(k+1)$:

$$(3m - 6k^2 - 5k) + 9k^2 + 20k + 12$$

Group terms by power of $k$:

$$= 3m + (-6k^2 + 9k^2) + (-5k + 20k) + 12$$

$$= 3m + 3k^2 + 15k + 12$$

Factor out 3 from the expression:

$$= 3 (m + k^2 + 5k + 4)$$

Since $m$ is an integer (from the Inductive Hypothesis) and $k$ is an integer, the term $(m + k^2 + 5k + 4)$ is also an integer.

Let $p = m + k^2 + 5k + 4$, where $p$ is an integer. Then we have:

$$(k+1)(k+2)(k+6) = 3p$$

This shows that $(k+1)(k+2)(k+6)$ is a multiple of 3, which means it is divisible by 3.

Thus, we have shown that $P(k+1)$ is true whenever $P(k)$ is true.

Conclusion:

Since $P(1)$ is true (Base Case) and $P(k+1)$ is true whenever $P(k)$ is true for $k \geq 1$ (Inductive Step), by the Principle of Mathematical Induction, the statement $P(n): n(n+1)(n+5)$ is divisible by 3 is true for all natural numbers $n \in \mathbb{N}$.

Given:

A statement $P(n)$ concerning the sum of a series where the terms are reciprocals of triangular numbers.

The $i$-th term of the series is $\frac{1}{1+2+\dots+i}$. The sum $1+2+\dots+i$ is the sum of the first $i$ natural numbers, which is given by the formula $\frac{i(i+1)}{2}$.

So, the $i$-th term is $\frac{1}{\frac{i(i+1)}{2}} = \frac{2}{i(i+1)}$.

The statement $P(n)$ can be written as:

$$P(n): \sum\limits_{i=1}^{n} \frac{2}{i(i+1)} = 1 + \frac{2}{2 \cdot 3} + \frac{2}{3 \cdot 4} + \dots + \frac{2}{n(n+1)} = \frac{2n}{n+1}$$

for all $n \in \mathbb{N}$. Note that for $i=1$, $\frac{2}{1(1+1)} = \frac{2}{2} = 1$, which matches the first term of the sum as stated in the question.

To Prove:

The statement $P(n)$ is true for all natural numbers $n$, using the Principle of Mathematical Induction.

Proof:

We prove the statement $P(n)$ using the Principle of Mathematical Induction. The principle involves two steps:

Step 1: Base Case

We verify if the statement is true for the smallest natural number, $n=1$.

For $n=1$, the left-hand side (LHS) of $P(n)$ is the first term of the sum as given in the question, which is $1$.

$$LHS = 1$$

The right-hand side (RHS) of $P(n)$ for $n=1$ is $\frac{2(1)}{1+1}$.

$$RHS = \frac{2(1)}{1+1} = \frac{2}{2} = 1$$

Since $LHS = RHS$, the statement $P(1)$ is true.

Step 2: Inductive Hypothesis

Assume that the statement $P(k)$ is true for some arbitrary positive integer $k$. That is, we assume:

Using the expanded form of the terms, this assumption is:

$$\sum\limits_{i=1}^{k} \frac{2}{i(i+1)} = \frac{2}{1 \cdot 2} + \frac{2}{2 \cdot 3} + \dots + \frac{2}{k(k+1)} = \frac{2k}{k+1}$$

Step 3: Inductive Step

We need to prove that the statement $P(k+1)$ is true whenever $P(k)$ is true. That is, we need to prove that:

$$1 + \frac{1}{1+2} + \dots + \frac{1}{1+2+\dots+k} + \frac{1}{1+2+\dots+(k+1)} = \frac{2(k+1)}{(k+1)+1}$$

$$1 + \frac{1}{1+2} + \dots + \frac{1}{1+2+\dots+k} + \frac{1}{\frac{(k+1)((k+1)+1)}{2}} = \frac{2(k+1)}{k+2}$$

$$1 + \frac{1}{1+2} + \dots + \frac{1}{1+2+\dots+k} + \frac{2}{(k+1)(k+2)} = \frac{2(k+1)}{k+2}$$

Consider the LHS of $P(k+1)$:

$$LHS = \left[1 + \frac{1}{1+2} + \dots + \frac{1}{1+2+\dots+k}\right] + \frac{2}{(k+1)(k+2)}$$

Using the Inductive Hypothesis from equation (i), we substitute $\frac{2k}{k+1}$ for the sum of the first $k$ terms:

$$LHS = \frac{2k}{k+1} + \frac{2}{(k+1)(k+2)}$$

To combine the fractions, find a common denominator, which is $(k+1)(k+2)$:

$$LHS = \frac{2k(k+2)}{(k+1)(k+2)} + \frac{2}{(k+1)(k+2)}$$

Combine the numerators:

$$LHS = \frac{2k(k+2) + 2}{(k+1)(k+2)}$$

Simplify the numerator:

$$2k(k+2) + 2 = 2k^2 + 4k + 2$$

Factor out 2 from the numerator:

$$2k^2 + 4k + 2 = 2(k^2 + 2k + 1)$$

Recognize the perfect square in the parenthesis:

$$k^2 + 2k + 1 = (k+1)^2$$

Substitute the factored numerator back into the LHS expression:

$$LHS = \frac{2(k+1)^2}{(k+1)(k+2)}$$

Assuming $k+1 \neq 0$ (which is true since $k \in \mathbb{N}$), we can cancel one factor of $(k+1)$ from the numerator and denominator:

$$LHS = \frac{2\cancel{(k+1)}^{\,1} (k+1)}{\cancel{(k+1)}_{\,1} (k+2)} = \frac{2(k+1)}{k+2}$$

This is the RHS of the statement $P(k+1)$.

Thus, we have shown that $LHS = RHS$, which means $P(k+1)$ is true whenever $P(k)$ is true.

Conclusion:

Since $P(1)$ is true (Base Case) and $P(k+1)$ is true whenever $P(k)$ is true (Inductive Step), by the Principle of Mathematical Induction, the statement $P(n): 1 + \frac{1}{1+2} + \dots + \frac{1}{1+2+\dots+n} = \frac{2n}{n+1}$ is true for all natural numbers $n \in \mathbb{N}$.

Given:

A statement $P(n)$ concerning an inequality involving $n$ and $x$.

$P(n): (1+x)^n \geq 1+nx$, for all $n \in \mathbb{N}$ and for a real number $x$ such that $x > -1$.

To Prove:

The statement $P(n)$ is true for all natural numbers $n$ and for all $x > -1$, using the Principle of Mathematical Induction.

Proof:

We prove the statement $P(n)$ using the Principle of Mathematical Induction on $n$. The principle involves two steps:

Step 1: Base Case

We verify if the statement is true for the smallest natural number, $n=1$.

For $n=1$, the left-hand side (LHS) of $P(n)$ is $(1+x)^1$.

$$LHS = (1+x)^1 = 1+x$$

The right-hand side (RHS) of $P(n)$ for $n=1$ is $1+1 \cdot x$.

$$RHS = 1+1 \cdot x = 1+x$$

Since $LHS = RHS$, we have $1+x \geq 1+x$. Thus, the inequality holds for $n=1$, and the statement $P(1)$ is true.

Step 2: Inductive Hypothesis

Assume that the statement $P(k)$ is true for some arbitrary positive integer $k \geq 1$. That is, for a given $x > -1$, we assume:

Step 3: Inductive Step

We need to prove that the statement $P(k+1)$ is true whenever $P(k)$ is true. That is, we need to prove that $(1+x)^{k+1} \geq 1+(k+1)x$.

Consider the LHS of $P(k+1)$:

$$LHS = (1+x)^{k+1}$$

We can rewrite $(1+x)^{k+1}$ as $(1+x)^k (1+x)$.

$$LHS = (1+x)^k (1+x)$$

From the Inductive Hypothesis (inequality i), we have $(1+x)^k \geq 1+kx$. Since $x > -1$, we have $1+x > 0$. Multiplying both sides of the inequality $(1+x)^k \geq 1+kx$ by the positive quantity $(1+x)$ preserves the direction of the inequality:

So,

$$(1+x)^{k+1} \geq (1+kx)(1+x)$$

Now, expand the right-hand side $(1+kx)(1+x)$:

$$(1+kx)(1+x) = 1 \cdot 1 + 1 \cdot x + kx \cdot 1 + kx \cdot x$$

$$= 1 + x + kx + kx^2$$

$$= 1 + (k+1)x + kx^2$$

Substituting this back into our inequality, we get:

$$ (1+x)^{k+1} \geq 1 + (k+1)x + kx^2 $$

We need to show that $(1+x)^{k+1} \geq 1+(k+1)x$. Comparing the expression $1 + (k+1)x + kx^2$ with $1+(k+1)x$, the difference is $kx^2$.

Since $k$ is a positive integer ($k \geq 1$), $k$ is positive ($k > 0$).

For any real number $x$, $x^2 \geq 0$.

Therefore, the product $kx^2$ is non-negative:

This means that $1 + (k+1)x + kx^2$ is greater than or equal to $1+(k+1)x$:

$$1 + (k+1)x + kx^2 \geq 1+(k+1)x$$

Combining the inequalities $(1+x)^{k+1} \geq 1 + (k+1)x + kx^2$ and $1 + (k+1)x + kx^2 \geq 1+(k+1)x$, by the transitive property of inequality, we conclude:

$$ (1+x)^{k+1} \geq 1+(k+1)x $$

This is the statement $P(k+1)$.

Thus, we have shown that $P(k+1)$ is true whenever $P(k)$ is true.

Conclusion:

Since $P(1)$ is true (Base Case) and $P(k+1)$ is true whenever $P(k)$ is true for $k \geq 1$ and $x > -1$ (Inductive Step), by the Principle of Mathematical Induction, the statement $P(n): (1+x)^n \geq 1+nx$ is true for all natural numbers $n \in \mathbb{N}$ and for all real numbers $x > -1$.

Given:

A statement $P(n)$ concerning an inequality.

$P(n): 2^n \geq n+1$, for all natural numbers $n \in \mathbb{N}$.

To Prove:

The statement $P(n)$ is true for all natural numbers $n$, using the Principle of Mathematical Induction.

Proof:

We prove the statement $P(n)$ using the Principle of Mathematical Induction. The principle involves two steps:

Step 1: Base Case

We verify if the statement is true for the smallest natural number, $n=1$.

For $n=1$, the left-hand side (LHS) of $P(n)$ is $2^1$.

$$LHS = 2^1 = 2$$

The right-hand side (RHS) of $P(n)$ for $n=1$ is $1+1$.

$$RHS = 1+1 = 2$$

Since $LHS = RHS$, we have $2 \geq 2$. Thus, the inequality holds for $n=1$, and the statement $P(1)$ is true.

Step 2: Inductive Hypothesis

Assume that the statement $P(k)$ is true for some arbitrary positive integer $k \geq 1$. That is, we assume:

Step 3: Inductive Step

We need to prove that the statement $P(k+1)$ is true whenever $P(k)$ is true. That is, we need to prove that $2^{k+1} \geq (k+1)+1$, which simplifies to $2^{k+1} \geq k+2$.

Consider the LHS of $P(k+1)$:

$$LHS = 2^{k+1}$$

We can rewrite $2^{k+1}$ as $2 \cdot 2^k$.

$$LHS = 2 \cdot 2^k$$

Using the Inductive Hypothesis from inequality (i), we know that $2^k \geq k+1$. Multiply both sides of this inequality by $2$ (which is a positive number, so the inequality direction is preserved):

So,

$$2^{k+1} \geq 2k + 2$$

We need to show that $2^{k+1} \geq k+2$. We have $2^{k+1} \geq 2k+2$. Now, we need to show that $2k+2 \geq k+2$ for $k \geq 1$.

Consider the difference $(2k+2) - (k+2)$:

$$(2k+2) - (k+2) = 2k + 2 - k - 2 = k$$

Since $k$ is a positive integer ($k \geq 1$), we have $k > 0$.

This implies:

Combining the inequalities $2^{k+1} \geq 2k+2$ and $2k+2 \geq k+2$, we can conclude that:

$$2^{k+1} \geq 2k+2 \geq k+2$$

By the transitive property of inequality, this gives us:

$$2^{k+1} \geq k+2$$

This is the RHS of the statement $P(k+1)$.

Thus, we have shown that $P(k+1)$ is true whenever $P(k)$ is true.

Conclusion:

Since $P(1)$ is true (Base Case) and $P(k+1)$ is true whenever $P(k)$ is true for $k \geq 1$ (Inductive Step), by the Principle of Mathematical Induction, the statement $P(n): 2^n \geq n+1$ is true for all natural numbers $n \in \mathbb{N}$.

Given:

A statement $P(n)$ concerning the divisibility of an expression by 9.

$P(n): 4^n + 15n - 1$ is divisible by 9, for all natural numbers $n \in \mathbb{N}$.

To Prove:

The statement $P(n)$ is true for all natural numbers $n$, using the Principle of Mathematical Induction.

Proof:

We prove the statement $P(n)$ using the Principle of Mathematical Induction. The principle involves two steps:

Step 1: Base Case

We verify if the statement is true for the smallest natural number, $n=1$.

For $n=1$, we evaluate the expression $4^1 + 15(1) - 1$:

$$4^1 + 15(1) - 1 = 4 + 15 - 1 = 19 - 1 = 18$$

We check if $18$ is divisible by 9.

$$18 \div 9 = 2$$

Since 18 is divisible by 9, the statement $P(1)$ is true.

Step 2: Inductive Hypothesis

Assume that the statement $P(k)$ is true for some arbitrary positive integer $k \geq 1$.

This means that $4^k + 15k - 1$ is divisible by 9. Therefore, we can write:

where $m$ is some integer.

From equation (i), we can express $4^k$ as:

Step 3: Inductive Step

We need to prove that the statement $P(k+1)$ is true whenever $P(k)$ is true. That is, we need to prove that $4^{k+1} + 15(k+1) - 1$ is divisible by 9.

Consider the expression for $P(k+1)$:

$$4^{k+1} + 15(k+1) - 1$$

$$= 4^{k+1} + 15k + 15 - 1$$

$$= 4^{k+1} + 15k + 14$$

We can rewrite $4^{k+1}$ as $4 \cdot 4^k$. Substitute the expression for $4^k$ from the Inductive Hypothesis ($4^k = 9m - 15k + 1$):

$$= 4 \cdot (9m - 15k + 1) + 15k + 14$$

Distribute the 4:

$$= 4 \cdot 9m - 4 \cdot 15k + 4 \cdot 1 + 15k + 14$$

$$= 36m - 60k + 4 + 15k + 14$$

Combine the terms with $k$ and the constant terms:

$$= 36m + (-60k + 15k) + (4 + 14)$$

$$= 36m - 45k + 18$$

Now, factor out 9 from the expression:

$$= 9 (4m - 5k + 2)$$

Since $m$ is an integer (from the Inductive Hypothesis) and $k$ is an integer, the term $(4m - 5k + 2)$ is also an integer.

Let $p = 4m - 5k + 2$, where $p$ is an integer. Then we have:

$$4^{k+1} + 15(k+1) - 1 = 9p$$

This shows that $4^{k+1} + 15(k+1) - 1$ is a multiple of 9, which means it is divisible by 9.

Thus, we have shown that $P(k+1)$ is true whenever $P(k)$ is true.

Conclusion:

Since $P(1)$ is true (Base Case) and $P(k+1)$ is true whenever $P(k)$ is true for $k \geq 1$ (Inductive Step), by the Principle of Mathematical Induction, the statement $P(n): 4^n + 15n - 1$ is divisible by 9 is true for all natural numbers $n \in \mathbb{N}$.

Given:

A statement $P(n)$ concerning the sum of the squares of the first $n$ odd natural numbers.

$$P(n): 1^2 + 3^2 + \dots + (2n-1)^2 = \frac{n(2n-1)(2n+1)}{3}$$

for all $n \in \mathbb{N}$.

To Prove:

The statement $P(n)$ is true for all natural numbers $n$, using the Principle of Mathematical Induction.

Proof:

We prove the statement $P(n)$ using the Principle of Mathematical Induction. The principle involves two steps:

Step 1: Base Case

We verify if the statement is true for the smallest natural number, $n=1$.

For $n=1$, the left-hand side (LHS) of $P(n)$ is the first term of the sum.

$$LHS = (2(1)-1)^2 = (2-1)^2 = 1^2 = 1$$

The right-hand side (RHS) of $P(n)$ for $n=1$ is $\frac{1(2(1)-1)(2(1)+1)}{3}$.

$$RHS = \frac{1(2-1)(2+1)}{3} = \frac{1(1)(3)}{3} = \frac{3}{3} = 1$$

Since $LHS = RHS$, the statement $P(1)$ is true.

Step 2: Inductive Hypothesis

Assume that the statement $P(k)$ is true for some arbitrary positive integer $k$. That is, we assume:

Step 3: Inductive Step

We need to prove that the statement $P(k+1)$ is true whenever $P(k)$ is true. That is, we need to prove that:

$$1^2 + 3^2 + \dots + (2k-1)^2 + (2(k+1)-1)^2 = \frac{(k+1)(2(k+1)-1)(2(k+1)+1)}{3}$$

The $(k+1)$-th term is $(2k+2-1)^2 = (2k+1)^2$.

The RHS for $n=k+1$ is $\frac{(k+1)(2k+1)(2k+3)}{3}$.

Consider the LHS of $P(k+1)$:

$$LHS = 1^2 + 3^2 + \dots + (2k-1)^2 + (2k+1)^2$$

We can group the first $k$ terms, which is the LHS of $P(k)$.

$$LHS = [1^2 + 3^2 + \dots + (2k-1)^2] + (2k+1)^2$$

Using the Inductive Hypothesis from equation (i), we substitute $\frac{k(2k-1)(2k+1)}{3}$ for the sum of the first $k$ terms:

$$LHS = \frac{k(2k-1)(2k+1)}{3} + (2k+1)^2$$

To combine the terms, find a common denominator:

$$LHS = \frac{k(2k-1)(2k+1)}{3} + \frac{3(2k+1)^2}{3}$$

Combine the numerators:

$$LHS = \frac{k(2k-1)(2k+1) + 3(2k+1)^2}{3}$$

Factor out the common term $(2k+1)$ from the numerator:

$$LHS = \frac{(2k+1) [k(2k-1) + 3(2k+1)]}{3}$$

Simplify the expression inside the square brackets:

$$k(2k-1) + 3(2k+1) = 2k^2 - k + 6k + 3 = 2k^2 + 5k + 3$$

Now, substitute this back into the expression for LHS:

$$LHS = \frac{(2k+1) (2k^2 + 5k + 3)}{3}$$

Factor the quadratic $2k^2 + 5k + 3$. We look for two numbers that multiply to $2 \times 3 = 6$ and add up to $5$. These numbers are $2$ and $3$.

$$2k^2 + 5k + 3 = 2k^2 + 2k + 3k + 3$$

$$= 2k(k+1) + 3(k+1)$$

$$= (2k+3)(k+1)$$

Substitute the factored quadratic back into the LHS expression:

$$LHS = \frac{(2k+1) (k+1)(2k+3)}{3}$$

Rearrange the factors to match the RHS of $P(k+1)$:

$$LHS = \frac{(k+1)(2k+1)(2k+3)}{3}$$

This is the RHS of the statement $P(k+1)$.

Thus, we have shown that $LHS = RHS$, which means $P(k+1)$ is true whenever $P(k)$ is true.

Conclusion:

Since $P(1)$ is true (Base Case) and $P(k+1)$ is true whenever $P(k)$ is true (Inductive Step), by the Principle of Mathematical Induction, the statement $P(n): 1^2 + 3^2 + \dots + (2n-1)^2 = \frac{n(2n-1)(2n+1)}{3}$ is true for all natural numbers $n \in \mathbb{N}$.

Given:

A statement $P(n)$ concerning the divisibility of the difference of powers of two numbers.

$P(n): a^n - b^n$ is divisible by $a-b$, for all natural numbers $n \in \mathbb{N}$, where $a$ and $b$ are real numbers such that $a \neq b$. (Assuming $a, b$ are such that $a-b$ is not zero, and the context implies integers or real numbers where divisibility means $\frac{a^n - b^n}{a-b}$ is an integer for integer $a,b$, or simply that $a^n - b^n = k(a-b)$ for some $k$).

To Prove:

The statement $P(n)$ is true for all natural numbers $n$, using the Principle of Mathematical Induction.

Proof:

We prove the statement $P(n)$ using the Principle of Mathematical Induction. The principle involves two steps:

Step 1: Base Case

We verify if the statement is true for the smallest natural number, $n=1$.

For $n=1$, we evaluate the expression $a^1 - b^1$:

$$a^1 - b^1 = a - b$$

We check if $a-b$ is divisible by $a-b$. Since $a \neq b$, $a-b$ is non-zero. Any non-zero number is divisible by itself.

$$\frac{a-b}{a-b} = 1$$

Since 1 is an integer, $a-b$ is divisible by $a-b$. Thus, the statement $P(1)$ is true.

Step 2: Inductive Hypothesis

Assume that the statement $P(k)$ is true for some arbitrary positive integer $k \geq 1$.

This means that $a^k - b^k$ is divisible by $a-b$. Therefore, we can write:

where $m$ is an integer (assuming $a, b$ are integers for standard definition of divisibility, or $m$ is some real number). From equation (i), we can express $a^k$ as:

Step 3: Inductive Step

We need to prove that the statement $P(k+1)$ is true whenever $P(k)$ is true. That is, we need to prove that $a^{k+1} - b^{k+1}$ is divisible by $a-b$.

Consider the expression for $P(k+1)$:

$$a^{k+1} - b^{k+1}$$

We can rewrite this expression and use the Inductive Hypothesis. Let's add and subtract the term $a^k b$:

$$a^{k+1} - b^{k+1} = a^{k+1} - a^k b + a^k b - b^{k+1}$$

Group the terms:

$$= (a^{k+1} - a^k b) + (a^k b - b^{k+1})$$

Factor out common terms from each group:

$$= a^k(a - b) + b(a^k - b^k)$$

Using the Inductive Hypothesis from equation (i), substitute $m(a-b)$ for $a^k - b^k$:

$$= a^k(a - b) + b(m(a-b))$$

Factor out the common term $(a-b)$:

$$= (a - b)(a^k + bm)$$

Since $a, b$ are numbers and $m$ is an integer (or a real number depending on the domain of $a, b$), the term $(a^k + bm)$ is also of the same type (integer if $a,b,m$ are integers). If $a,b$ are integers and $a-b \neq 0$, then $a^k+bm$ is an integer because $a^k-b^k$ is divisible by $a-b$.

Let $p = a^k + bm$. Since $a, b, k, m$ are such that $p$ is an appropriate number (integer if divisibility in the standard sense is meant), we have:

$$a^{k+1} - b^{k+1} = p(a-b)$$

This shows that $a^{k+1} - b^{k+1}$ is a multiple of $a-b$, which means it is divisible by $a-b$.

Thus, we have shown that $P(k+1)$ is true whenever $P(k)$ is true.

Conclusion:

Since $P(1)$ is true (Base Case) and $P(k+1)$ is true whenever $P(k)$ is true for $k \geq 1$ (Inductive Step), by the Principle of Mathematical Induction, the statement $P(n): a^n - b^n$ is divisible by $a-b$ is true for all natural numbers $n \in \mathbb{N}$, where $a \neq b$.

Given:

A statement $P(n)$ concerning an inequality.

$P(n): n^2 < 2^n$, for all integers $n \geq 5$.

To Prove:

The statement $P(n)$ is true for all integers $n \geq 5$, using the Principle of Mathematical Induction.

Proof:

We prove the statement $P(n)$ using the Principle of Mathematical Induction. The principle involves two steps:

Step 1: Base Case

We verify if the statement is true for the smallest integer in the specified range, $n=5$.

For $n=5$, the left-hand side (LHS) of $P(n)$ is $5^2$.

$$LHS = 5^2 = 25$$

The right-hand side (RHS) of $P(n)$ for $n=5$ is $2^5$.

$$RHS = 2^5 = 32$$

Since $25 < 32$, which is $LHS < RHS$, the statement $P(5)$ is true.

Step 2: Inductive Hypothesis

Assume that the statement $P(k)$ is true for some arbitrary integer $k \geq 5$. That is, we assume:

Step 3: Inductive Step

We need to prove that the statement $P(k+1)$ is true whenever $P(k)$ is true. That is, we need to prove that $(k+1)^2 < 2^{k+1}$.

Consider the LHS of $P(k+1)$:

$$LHS = (k+1)^2$$

Expand the expression:

$$LHS = k^2 + 2k + 1$$

From the Inductive Hypothesis (inequality i), we know that $k^2 < 2^k$. Substitute this into the expression:

$$(k+1)^2 < 2^k + 2k + 1$$

Now, we need to show that $2^k + 2k + 1 < 2^{k+1}$. This is equivalent to showing $2^k + 2k + 1 < 2 \cdot 2^k$, or $2k + 1 < 2 \cdot 2^k - 2^k$, which simplifies to $2k + 1 < 2^k$.

Let's prove that $2k+1 < 2^k$ for all integers $k \geq 5$. We can do this by examining the values or by a small inductive argument if necessary. For $k=5$, $2(5)+1=11$ and $2^5=32$, $11 < 32$. The difference $2^k - (2k+1)$ is increasing for $k \geq 2$. Since it's positive for $k=5$, it's positive for all $k \geq 5$. Therefore, $2k+1 < 2^k$ for all integers $k \geq 5$.

Using the fact that $2k+1 < 2^k$ for $k \geq 5$, we can continue from the inequality $(k+1)^2 < 2^k + 2k + 1$:

$$(k+1)^2 < 2^k + (2k + 1)$$

Since $2k+1 < 2^k$ for $k \geq 5$, we can replace $(2k+1)$ with $2^k$ in the inequality to get a larger value on the RHS:

$$(k+1)^2 < 2^k + 2^k$$

$$(k+1)^2 < 2 \cdot 2^k$$

$$(k+1)^2 < 2^{k+1}$$

This is the statement $P(k+1)$.

Thus, we have shown that $P(k+1)$ is true whenever $P(k)$ is true for $k \geq 5$.

Conclusion:

Since $P(5)$ is true (Base Case) and $P(k+1)$ is true whenever $P(k)$ is true for $k \geq 5$ (Inductive Step), by the Principle of Mathematical Induction, the statement $P(n): n^2 < 2^n$ is true for all integers $n \geq 5$.

Given:

A statement $P(n)$ concerning the divisibility of an expression by 11.

$P(n): 10^{2n-1} + 1$ is divisible by 11, for all natural numbers $n \in \mathbb{N}$.

To Prove:

The statement $P(n)$ is true for all natural numbers $n$, using the Principle of Mathematical Induction.

Proof:

We prove the statement $P(n)$ using the Principle of Mathematical Induction. The principle involves two steps:

Step 1: Base Case

We verify if the statement is true for the smallest natural number, $n=1$.

For $n=1$, we evaluate the expression $10^{2(1)-1} + 1$:

$$10^{2(1)-1} + 1 = 10^{2-1} + 1$$

$$= 10^1 + 1$$

$$= 10 + 1 = 11$$

We check if $11$ is divisible by 11.

$$11 \div 11 = 1$$

Since 11 is divisible by 11, the statement $P(1)$ is true.

Step 2: Inductive Hypothesis

Assume that the statement $P(k)$ is true for some arbitrary positive integer $k \geq 1$.

This means that $10^{2k-1} + 1$ is divisible by 11. Therefore, we can write:

where $m$ is some integer.

From equation (i), we can express $10^{2k-1}$ as:

Step 3: Inductive Step

We need to prove that the statement $P(k+1)$ is true whenever $P(k)$ is true. That is, we need to prove that $10^{2(k+1)-1} + 1$ is divisible by 11.

Consider the expression for $P(k+1)$:

$$10^{2(k+1)-1} + 1$$

Simplify the exponent:

$$= 10^{2k+2-1} + 1$$

$$= 10^{2k+1} + 1$$

Rewrite $10^{2k+1}$ using properties of exponents:

$$= 10^{2k-1+2} + 1$$

$$= 10^{2k-1} \cdot 10^2 + 1$$

$$= 100 \cdot 10^{2k-1} + 1$$

Using the Inductive Hypothesis, substitute $10^{2k-1} = 11m - 1$ into the expression:

$$= 100 \cdot (11m - 1) + 1$$

Distribute the 100:

$$= 100 \cdot 11m - 100 \cdot 1 + 1$$

$$= 1100m - 100 + 1$$

Combine the constant terms:

$$= 1100m - 99$$

Now, factor out 11 from the expression:

$$= 11 (100m - 9)$$

Since $m$ is an integer (from the Inductive Hypothesis), the term $(100m - 9)$ is also an integer.

Let $p = 100m - 9$, where $p$ is an integer. Then we have:

$$10^{2(k+1)-1} + 1 = 11p$$

This shows that $10^{2(k+1)-1} + 1$ is a multiple of 11, which means it is divisible by 11.

Thus, we have shown that $P(k+1)$ is true whenever $P(k)$ is true.

Conclusion:

Since $P(1)$ is true (Base Case) and $P(k+1)$ is true whenever $P(k)$ is true for $k \geq 1$ (Inductive Step), by the Principle of Mathematical Induction, the statement $P(n): 10^{2n-1} + 1$ is divisible by 11 is true for all natural numbers $n \in \mathbb{N}$.

Given:

A statement $P(n)$ concerning the sum of the first $n$ terms of a geometric series with first term $a=1$ and common ratio $r=2$.

$$P(n): 1+2+2^2+\dots+2^{n-1} = 2^n-1$$

for all $n \in \mathbb{N}$.

To Prove:

The statement $P(n)$ is true for all natural numbers $n$, using the Principle of Mathematical Induction.

Proof:

We prove the statement $P(n)$ using the Principle of Mathematical Induction. The principle involves two steps:

Step 1: Base Case

We verify if the statement is true for the smallest natural number, $n=1$.

For $n=1$, the left-hand side (LHS) of $P(n)$ is the first term of the sum, which is $2^{1-1} = 2^0 = 1$.

$$LHS = 1$$

The right-hand side (RHS) of $P(n)$ for $n=1$ is $2^1-1$.

$$RHS = 2^1-1 = 2-1 = 1$$

Since $LHS = RHS$, the statement $P(1)$ is true.

Step 2: Inductive Hypothesis

Assume that the statement $P(k)$ is true for some arbitrary positive integer $k \geq 1$. That is, we assume:

Step 3: Inductive Step

We need to prove that the statement $P(k+1)$ is true whenever $P(k)$ is true. That is, we need to prove that:

$$1+2+2^2+\dots+2^{k-1}+2^{(k+1)-1} = 2^{(k+1)}-1$$

$$1+2+2^2+\dots+2^{k-1}+2^k = 2^{k+1}-1$$

Consider the LHS of $P(k+1)$:

$$LHS = 1+2+2^2+\dots+2^{k-1}+2^k$$

We can group the first $k$ terms, which is the LHS of $P(k)$.

$$LHS = [1+2+2^2+\dots+2^{k-1}] + 2^k$$

Using the Inductive Hypothesis from equation (i), we substitute $2^k-1$ for the sum of the first $k$ terms:

$$LHS = (2^k-1) + 2^k$$

Combine the terms:

$$LHS = 2^k + 2^k - 1$$

$$LHS = 2 \cdot 2^k - 1$$

Using the property of exponents $a^m \cdot a^n = a^{m+n}$:

$$LHS = 2^{1+k} - 1$$

$$LHS = 2^{k+1} - 1$$

This is the RHS of the statement $P(k+1)$.

Thus, we have shown that $LHS = RHS$, which means $P(k+1)$ is true whenever $P(k)$ is true.

Conclusion:

Since $P(1)$ is true (Base Case) and $P(k+1)$ is true whenever $P(k)$ is true (Inductive Step), by the Principle of Mathematical Induction, the statement $P(n): 1+2+2^2+\dots+2^{n-1} = 2^n-1$ is true for all natural numbers $n \in \mathbb{N}$.

Given:

A statement $P(n)$ concerning the divisibility of the product of three consecutive positive integers by 6.

The product of three consecutive positive integers starting with $n$ is $n(n+1)(n+2)$.

$P(n): n(n+1)(n+2)$ is divisible by 6, for all natural numbers $n \in \mathbb{N}$.

To Prove:

The statement $P(n)$ is true for all natural numbers $n$, using the Principle of Mathematical Induction.

Proof:

We prove the statement $P(n)$ using the Principle of Mathematical Induction. The principle involves two steps:

Step 1: Base Case

We verify if the statement is true for the smallest natural number, $n=1$.

For $n=1$, we evaluate the product $1(1+1)(1+2)$:

$$1(1+1)(1+2) = 1 \cdot 2 \cdot 3 = 6$$

We check if $6$ is divisible by 6.

$$6 \div 6 = 1$$

Since 6 is divisible by 6, the statement $P(1)$ is true.

Step 2: Inductive Hypothesis

Assume that the statement $P(k)$ is true for some arbitrary positive integer $k \geq 1$.

This means that $k(k+1)(k+2)$ is divisible by 6. Therefore, we can write:

where $m$ is some integer.

Step 3: Inductive Step

We need to prove that the statement $P(k+1)$ is true whenever $P(k)$ is true. That is, we need to prove that $(k+1)((k+1)+1)((k+1)+2)$ is divisible by 6.

Consider the expression for $P(k+1)$:

$$(k+1)((k+1)+1)((k+1)+2) = (k+1)(k+2)(k+3)$$

We can rewrite this expression by expanding it or by relating it to the inductive hypothesis. Let's relate it to $k(k+1)(k+2)$.

$$(k+1)(k+2)(k+3) = (k+1)(k+2) \cdot k + (k+1)(k+2) \cdot 3$$

$$= k(k+1)(k+2) + 3(k+1)(k+2)$$

From the Inductive Hypothesis (equation i), we know that $k(k+1)(k+2)$ is divisible by 6. So, $k(k+1)(k+2) = 6m$ for some integer $m$.

Thus, the expression becomes:

$$(k+1)(k+2)(k+3) = 6m + 3(k+1)(k+2)$$

To show that this entire expression is divisible by 6, we need to show that $3(k+1)(k+2)$ is also divisible by 6.

Consider the term $(k+1)(k+2)$. This is the product of two consecutive integers.

The product of any two consecutive integers is always divisible by 2, because one of the two integers must be even.

So, $(k+1)(k+2)$ is divisible by 2. We can write $(k+1)(k+2) = 2j$ for some integer $j$.

Now substitute this into the term $3(k+1)(k+2)$:

$$3(k+1)(k+2) = 3(2j) = 6j$$

Since $j$ is an integer, $6j$ is a multiple of 6, and thus, $3(k+1)(k+2)$ is divisible by 6.

Now, we return to the expression for $P(k+1)$:

$$(k+1)(k+2)(k+3) = k(k+1)(k+2) + 3(k+1)(k+2)$$

We know that $k(k+1)(k+2)$ is divisible by 6 (from $P(k)$) and $3(k+1)(k+2)$ is divisible by 6 (as shown above).

The sum of two integers that are divisible by 6 is also divisible by 6.

Let $k(k+1)(k+2) = 6m$ and $3(k+1)(k+2) = 6j$.

Then $(k+1)(k+2)(k+3) = 6m + 6j = 6(m+j)$.

Since $m$ and $j$ are integers, $(m+j)$ is an integer.

Therefore, $(k+1)(k+2)(k+3)$ is divisible by 6.

Thus, we have shown that $P(k+1)$ is true whenever $P(k)$ is true.

Conclusion:

Since $P(1)$ is true (Base Case) and $P(k+1)$ is true whenever $P(k)$ is true for $k \geq 1$ (Inductive Step), by the Principle of Mathematical Induction, the statement $P(n): n(n+1)(n+2)$ is divisible by 6 is true for all natural numbers $n \in \mathbb{N}$.

Given:

A statement $P(n)$ concerning the sum of a series.

$$P(n): \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \dots + \frac{1}{(2n+1)(2n+3)} = \frac{n}{3(2n+3)}$$

for all $n \in \mathbb{N}$.

To Prove:

The statement $P(n)$ is true for all natural numbers $n$, using the Principle of Mathematical Induction.

Proof:

We prove the statement $P(n)$ using the Principle of Mathematical Induction. The principle involves two steps:

Step 1: Base Case

We verify if the statement is true for the smallest natural number, $n=1$.

For $n=1$, the left-hand side (LHS) of $P(n)$ is the first term of the sum.

$$LHS = \frac{1}{3 \cdot (2(1)+3)} = \frac{1}{3 \cdot (2+3)} = \frac{1}{3 \cdot 5} = \frac{1}{15}$$

The right-hand side (RHS) of $P(n)$ for $n=1$ is $\frac{1}{3(2(1)+3)}$.

$$RHS = \frac{1}{3(2+3)} = \frac{1}{3(5)} = \frac{1}{15}$$

Since $LHS = RHS$, the statement $P(1)$ is true.

Step 2: Inductive Hypothesis

Assume that the statement $P(k)$ is true for some arbitrary positive integer $k$. That is, we assume:

Step 3: Inductive Step

We need to prove that the statement $P(k+1)$ is true whenever $P(k)$ is true. That is, we need to prove that:

$$\frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \dots + \frac{1}{(2k+1)(2k+3)} + \frac{1}{(2(k+1)+1)(2(k+1)+3)} = \frac{k+1}{3(2(k+1)+3)}$$

The $(k+1)$-th term of the series is $\frac{1}{(2k+2+1)(2k+2+3)} = \frac{1}{(2k+3)(2k+5)}$.

The RHS for $n=k+1$ is $\frac{k+1}{3(2k+2+3)} = \frac{k+1}{3(2k+5)}$.

Consider the LHS of $P(k+1)$:

$$LHS = \left[\frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \dots + \frac{1}{(2k+1)(2k+3)}\right] + \frac{1}{(2k+3)(2k+5)}$$

Using the Inductive Hypothesis from equation (i), we substitute $\frac{k}{3(2k+3)}$ for the sum of the first $k$ terms:

$$LHS = \frac{k}{3(2k+3)} + \frac{1}{(2k+3)(2k+5)}$$

To combine the fractions, find a common denominator, which is $3(2k+3)(2k+5)$:

$$LHS = \frac{k(2k+5)}{3(2k+3)(2k+5)} + \frac{3}{3(2k+3)(2k+5)}$$

Combine the numerators:

$$LHS = \frac{k(2k+5) + 3}{3(2k+3)(2k+5)}$$

Simplify the numerator:

$$k(2k+5) + 3 = 2k^2 + 5k + 3$$

Factor the quadratic expression $2k^2 + 5k + 3$. We look for two numbers that multiply to $2 \times 3 = 6$ and add up to $5$. These numbers are $2$ and $3$.

$$2k^2 + 5k + 3 = 2k^2 + 2k + 3k + 3$$

$$= 2k(k+1) + 3(k+1)$$

$$= (k+1)(2k+3)$$

Substitute the factored numerator back into the LHS expression:

$$LHS = \frac{(k+1)(2k+3)}{3(2k+3)(2k+5)}$$

Assuming $2k+3 \neq 0$ (which is true since $k \in \mathbb{N}$), we can cancel the factor $(2k+3)$ from the numerator and denominator:

$$LHS = \frac{(k+1)\cancel{(2k+3)}^{\,1}}{3\cancel{(2k+3)}_{\,1}(2k+5)} = \frac{k+1}{3(2k+5)}$$

This is the RHS of the statement $P(k+1)$.

Thus, we have shown that $LHS = RHS$, which means $P(k+1)$ is true whenever $P(k)$ is true.

Conclusion:

Since $P(1)$ is true (Base Case) and $P(k+1)$ is true whenever $P(k)$ is true (Inductive Step), by the Principle of Mathematical Induction, the statement $P(n): \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \dots + \frac{1}{(2n+1)(2n+3)} = \frac{n}{3(2n+3)}$ is true for all natural numbers $n \in \mathbb{N}$.

Given:

A statement $P(n)$ concerning the sum of the product of three consecutive natural numbers.

$$P(n): \sum\limits_{i=1}^{n} i(i+1)(i+2) = 1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + \dots + n(n+1)(n+2) = \frac{n(n+1)(n+2)(n+3)}{4}$$

for all $n \in \mathbb{N}$.

To Prove:

The statement $P(n)$ is true for all natural numbers $n$, using the Principle of Mathematical Induction.

Proof:

We prove the statement $P(n)$ using the Principle of Mathematical Induction. The principle involves two steps:

Step 1: Base Case

We verify if the statement is true for the smallest natural number, $n=1$.

For $n=1$, the left-hand side (LHS) of $P(n)$ is the first term of the sum.

$$LHS = 1 \cdot (1+1) \cdot (1+2) = 1 \cdot 2 \cdot 3 = 6$$

The right-hand side (RHS) of $P(n)$ for $n=1$ is $\frac{1(1+1)(1+2)(1+3)}{4}$.

$$RHS = \frac{1(2)(3)(4)}{4} = \frac{24}{4} = 6$$

Since $LHS = RHS$, the statement $P(1)$ is true.

Step 2: Inductive Hypothesis

Assume that the statement $P(k)$ is true for some arbitrary positive integer $k$. That is, we assume:

Step 3: Inductive Step

We need to prove that the statement $P(k+1)$ is true whenever $P(k)$ is true. That is, we need to prove that:

$$1 \cdot 2 \cdot 3 + \dots + k(k+1)(k+2) + (k+1)((k+1)+1)((k+1)+2) = \frac{(k+1)((k+1)+1)((k+1)+2)((k+1)+3)}{4}$$

$$1 \cdot 2 \cdot 3 + \dots + k(k+1)(k+2) + (k+1)(k+2)(k+3) = \frac{(k+1)(k+2)(k+3)(k+4)}{4}$$

Consider the LHS of $P(k+1)$:

$$LHS = 1 \cdot 2 \cdot 3 + \dots + k(k+1)(k+2) + (k+1)(k+2)(k+3)$$

We can group the first $k$ terms, which is the LHS of $P(k)$.

$$LHS = [1 \cdot 2 \cdot 3 + \dots + k(k+1)(k+2)] + (k+1)(k+2)(k+3)$$

Using the Inductive Hypothesis from equation (i), we substitute $\frac{k(k+1)(k+2)(k+3)}{4}$ for the sum of the first $k$ terms:

$$LHS = \frac{k(k+1)(k+2)(k+3)}{4} + (k+1)(k+2)(k+3)$$

To combine the terms, find a common denominator:

$$LHS = \frac{k(k+1)(k+2)(k+3)}{4} + \frac{4(k+1)(k+2)(k+3)}{4}$$

Combine the numerators:

$$LHS = \frac{k(k+1)(k+2)(k+3) + 4(k+1)(k+2)(k+3)}{4}$$

Factor out the common term $(k+1)(k+2)(k+3)$ from the numerator:

$$LHS = \frac{(k+1)(k+2)(k+3) [k + 4]}{4}$$

Rearrange the terms in the numerator:

$$LHS = \frac{(k+1)(k+2)(k+3)(k+4)}{4}$$

This is the RHS of the statement $P(k+1)$.

Thus, we have shown that $LHS = RHS$, which means $P(k+1)$ is true whenever $P(k)$ is true.

Conclusion:

Since $P(1)$ is true (Base Case) and $P(k+1)$ is true whenever $P(k)$ is true (Inductive Step), by the Principle of Mathematical Induction, the statement $P(n): 1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + \dots + n(n+1)(n+2) = \frac{n(n+1)(n+2)(n+3)}{4}$ is true for all natural numbers $n \in \mathbb{N}$.

Given:

A statement $P(n)$ concerning the divisibility of an expression by 24.

$P(n): 2 \cdot 7^n + 3 \cdot 5^n - 5$ is divisible by 24, for all natural numbers $n \in \mathbb{N}$.

To Prove:

The statement $P(n)$ is true for all natural numbers $n$, using the Principle of Mathematical Induction.

Proof:

We prove the statement $P(n)$ using the Principle of Mathematical Induction. The principle involves two steps:

Step 1: Base Case

We verify if the statement is true for the smallest natural number, $n=1$.

For $n=1$, we evaluate the expression $2 \cdot 7^1 + 3 \cdot 5^1 - 5$:

$$2 \cdot 7^1 + 3 \cdot 5^1 - 5 = 2 \cdot 7 + 3 \cdot 5 - 5$$

$$= 14 + 15 - 5$$

$$= 29 - 5 = 24$$

We check if $24$ is divisible by 24.

$$24 \div 24 = 1$$

Since 24 is divisible by 24, the statement $P(1)$ is true.

Step 2: Inductive Hypothesis

Assume that the statement $P(k)$ is true for some arbitrary positive integer $k \geq 1$.

This means that $2 \cdot 7^k + 3 \cdot 5^k - 5$ is divisible by 24. Therefore, we can write:

where $m$ is some integer.

Step 3: Inductive Step

We need to prove that the statement $P(k+1)$ is true whenever $P(k)$ is true. That is, we need to prove that $2 \cdot 7^{k+1} + 3 \cdot 5^{k+1} - 5$ is divisible by 24.

Consider the expression for $P(k+1)$:

$$E(k+1) = 2 \cdot 7^{k+1} + 3 \cdot 5^{k+1} - 5$$

We can rewrite the terms with exponents:

$$E(k+1) = 2 \cdot 7 \cdot 7^k + 3 \cdot 5 \cdot 5^k - 5$$

$$E(k+1) = 14 \cdot 7^k + 15 \cdot 5^k - 5$$

We want to relate this to the expression from the Inductive Hypothesis, $E(k) = 2 \cdot 7^k + 3 \cdot 5^k - 5$. Consider the difference $E(k+1) - 7 \cdot E(k)$:

$$E(k+1) - 7 \cdot E(k) = (14 \cdot 7^k + 15 \cdot 5^k - 5) - 7 \cdot (2 \cdot 7^k + 3 \cdot 5^k - 5)$$

$$= 14 \cdot 7^k + 15 \cdot 5^k - 5 - (14 \cdot 7^k + 21 \cdot 5^k - 35)$$

$$= 14 \cdot 7^k + 15 \cdot 5^k - 5 - 14 \cdot 7^k - 21 \cdot 5^k + 35$$

Combine like terms:

$$= (14 \cdot 7^k - 14 \cdot 7^k) + (15 \cdot 5^k - 21 \cdot 5^k) + (-5 + 35)$$

$$= 0 - 6 \cdot 5^k + 30$$

So, we have the relationship:

$$E(k+1) - 7 \cdot E(k) = -6 \cdot 5^k + 30$$

Rearrange to express $E(k+1)$:

$$E(k+1) = 7 \cdot E(k) - 6 \cdot 5^k + 30$$

From the Inductive Hypothesis (equation i), $E(k) = 24m$ for some integer $m$. Substitute this into the equation:

$$E(k+1) = 7 \cdot (24m) - 6 \cdot 5^k + 30$$

$$E(k+1) = 168m - 6(5^k - 5)$$

Now, we need to show that $-6(5^k - 5)$ is divisible by 24, which is equivalent to showing that $5^k - 5$ is divisible by 4 for all $k \geq 1$.

Let's prove that $5^k - 5$ is divisible by 4 for all $k \geq 1$ by considering the property of modular arithmetic. For any integer $k \geq 1$, $5 \equiv 1 \pmod{4}$.

Raising both sides to the power of $k$, we get:

$$5^k \equiv 1^k \pmod{4}$$

$$5^k \equiv 1 \pmod{4}$$

This means $5^k - 1$ is divisible by 4.

Consider $5^k - 5 = (5^k - 1) - 4$. Since $5^k - 1$ is divisible by 4 and 4 is divisible by 4, their difference $(5^k - 1) - 4$ is also divisible by 4.

Alternatively, using a small induction on $5^k - 5$ being divisible by 4:

Base case $k=1$: $5^1 - 5 = 0$, which is divisible by 4.

Assume $5^j - 5$ is divisible by 4 for some $j \geq 1$. So $5^j - 5 = 4q$ for some integer $q$.

Consider $5^{j+1} - 5 = 5 \cdot 5^j - 5 = 5(5^j) - 5$. We can write $5^j = 4q + 5$ from the assumption.

$$5(4q + 5) - 5 = 20q + 25 - 5 = 20q + 20 = 4(5q + 5)$$

Since $5q+5$ is an integer, $4(5q+5)$ is divisible by 4. Thus $5^{k}-5$ is divisible by 4 for all $k \geq 1$.

So, we can write $5^k - 5 = 4q$ for some integer $q$. Substitute this back into the expression for $E(k+1)$:

$$E(k+1) = 168m - 6(4q)$$

$$E(k+1) = 168m - 24q$$

Factor out 24:

$$E(k+1) = 24(7m - q)$$

Since $m$ and $q$ are integers, $7m - q$ is an integer. Therefore, $E(k+1)$ is a multiple of 24, which means it is divisible by 24.

Thus, we have shown that $P(k+1)$ is true whenever $P(k)$ is true.

Conclusion:

Since $P(1)$ is true (Base Case) and $P(k+1)$ is true whenever $P(k)$ is true for $k \geq 1$ (Inductive Step), by the Principle of Mathematical Induction, the statement $P(n): 2 \cdot 7^n + 3 \cdot 5^n - 5$ is divisible by 24 is true for all natural numbers $n \in \mathbb{N}$.

Given:

A statement $P(n)$ concerning the relationship between the sum of cubes of the first $n$ natural numbers and the square of the sum of the first $n$ natural numbers.

We know that the sum of the first $n$ natural numbers is $1+2+\dots+n = \frac{n(n+1)}{2}$.

So, the statement can be written as:

$$P(n): 1^3 + 2^3 + \dots + n^3 = \left(\frac{n(n+1)}{2}\right)^2$$

for all $n \in \mathbb{N}$.

To Prove:

The statement $P(n)$ is true for all natural numbers $n$, using the Principle of Mathematical Induction.

Proof:

We prove the statement $P(n)$ using the Principle of Mathematical Induction. The principle involves two steps:

Step 1: Base Case

We verify if the statement is true for the smallest natural number, $n=1$.

For $n=1$, the left-hand side (LHS) of $P(n)$ is the first term of the sum of cubes.

$$LHS = 1^3 = 1$$

The right-hand side (RHS) of $P(n)$ for $n=1$ is $(1)^2$ or $\left(\frac{1(1+1)}{2}\right)^2$.

$$RHS = \left(\frac{1(1+1)}{2}\right)^2 = \left(\frac{2}{2}\right)^2 = 1^2 = 1$$

Since $LHS = RHS$, the statement $P(1)$ is true.

Step 2: Inductive Hypothesis

Assume that the statement $P(k)$ is true for some arbitrary positive integer $k$. That is, we assume:

Step 3: Inductive Step

We need to prove that the statement $P(k+1)$ is true whenever $P(k)$ is true. That is, we need to prove that:

$$1^3 + 2^3 + \dots + k^3 + (k+1)^3 = \left(\frac{(k+1)((k+1)+1)}{2}\right)^2$$

$$1^3 + 2^3 + \dots + k^3 + (k+1)^3 = \left(\frac{(k+1)(k+2)}{2}\right)^2$$

Consider the LHS of $P(k+1)$:

$$LHS = 1^3 + 2^3 + \dots + k^3 + (k+1)^3$$

We can group the first $k$ terms, which is the LHS of $P(k)$.

$$LHS = [1^3 + 2^3 + \dots + k^3] + (k+1)^3$$

Using the Inductive Hypothesis from equation (i), we substitute $\left(\frac{k(k+1)}{2}\right)^2$ for the sum of the first $k$ terms:

$$LHS = \left(\frac{k(k+1)}{2}\right)^2 + (k+1)^3$$

$$LHS = \frac{k^2(k+1)^2}{4} + (k+1)^3$$

To combine the terms, find a common denominator:

$$LHS = \frac{k^2(k+1)^2}{4} + \frac{4(k+1)^3}{4}$$

Combine the numerators:

$$LHS = \frac{k^2(k+1)^2 + 4(k+1)^3}{4}$$

Factor out the common term $(k+1)^2$ from the numerator:

$$LHS = \frac{(k+1)^2 [k^2 + 4(k+1)]}{4}$$

Simplify the expression inside the square brackets:

$$k^2 + 4(k+1) = k^2 + 4k + 4$$

This is a perfect square trinomial, which can be factored as $(k+2)^2$.

$$k^2 + 4k + 4 = (k+2)^2$$

Now, substitute this back into the expression for LHS:

$$LHS = \frac{(k+1)^2 (k+2)^2}{4}$$

This expression can be written as:

$$LHS = \left(\frac{(k+1)(k+2)}{2}\right)^2$$

This is the RHS of the statement $P(k+1)$.

Thus, we have shown that $LHS = RHS$, which means $P(k+1)$ is true whenever $P(k)$ is true.

Conclusion:

Since $P(1)$ is true (Base Case) and $P(k+1)$ is true whenever $P(k)$ is true (Inductive Step), by the Principle of Mathematical Induction, the statement $P(n): 1^3 + 2^3 + \dots + n^3 = (1+2+\dots+n)^2$ is true for all natural numbers $n \in \mathbb{N}$.

Given:

A statement $P(n)$ concerning an inequality involving factorial and powers of 2.

$P(n): n! > 2^n$, for all integers $n \geq 4$.

To Prove:

The statement $P(n)$ is true for all integers $n \geq 4$, using the Principle of Mathematical Induction.

Proof:

We prove the statement $P(n)$ using the Principle of Mathematical Induction. The principle involves two steps:

Step 1: Base Case

We verify if the statement is true for the smallest integer in the specified range, $n=4$.

For $n=4$, the left-hand side (LHS) of $P(n)$ is $4!$.

$$LHS = 4! = 4 \times 3 \times 2 \times 1 = 24$$

The right-hand side (RHS) of $P(n)$ for $n=4$ is $2^4$.

$$RHS = 2^4 = 2 \times 2 \times 2 \times 2 = 16$$

Since $24 > 16$, which is $LHS > RHS$, the statement $P(4)$ is true.

Step 2: Inductive Hypothesis

Assume that the statement $P(k)$ is true for some arbitrary integer $k \geq 4$. That is, we assume:

Step 3: Inductive Step

We need to prove that the statement $P(k+1)$ is true whenever $P(k)$ is true. That is, we need to prove that $(k+1)! > 2^{k+1}$.

Consider the LHS of $P(k+1)$:

$$LHS = (k+1)!$$

Using the definition of factorial, we can write $(k+1)!$ as $(k+1) \cdot k!$.

$$LHS = (k+1) \cdot k!$$

Using the Inductive Hypothesis from inequality (i), we know that $k! > 2^k$. Multiply both sides of this inequality by $(k+1)$. Since $k \geq 4$, $k+1 \geq 5$, so $(k+1)$ is a positive number. Multiplying by a positive number preserves the direction of the inequality:

So,

Now, we need to show that $(k+1) \cdot 2^k \geq 2^{k+1}$. We can rewrite $2^{k+1}$ as $2 \cdot 2^k$. So, we need to show $(k+1) \cdot 2^k \geq 2 \cdot 2^k$. This is equivalent to showing that $k+1 \geq 2$.

Since $k$ is an integer and $k \geq 4$, we have $k+1 \geq 4+1 = 5$.

Since $k+1 \geq 5$ and $5 > 2$, it follows that $k+1 > 2$. More precisely, $k+1 \geq 2$ holds true for all $k \geq 1$. As our base case is $k=4$, this condition is met.

Multiplying both sides of this inequality by $2^k$ (which is positive since $k \geq 4$), we get:

$$ (k+1) \cdot 2^k \geq 2^{k+1} $$

Combining inequality (ii) and inequality (iii), we have:

$$ (k+1)! > (k+1) \cdot 2^k \geq 2^{k+1} $$

By the transitive property of inequality, this gives us:

$$ (k+1)! > 2^{k+1} $$

This is the statement $P(k+1)$.

Thus, we have shown that $P(k+1)$ is true whenever $P(k)$ is true for $k \geq 4$.

Conclusion:

Since $P(4)$ is true (Base Case) and $P(k+1)$ is true whenever $P(k)$ is true for $k \geq 4$ (Inductive Step), by the Principle of Mathematical Induction, the statement $P(n): n! > 2^n$ is true for all integers $n \geq 4$.

Given:

A statement $P(n)$ concerning the divisibility of an expression by 27.

$P(n): 41^n - 14^n$ is divisible by 27, for all natural numbers $n \in \mathbb{N}$.

To Prove:

The statement $P(n)$ is true for all natural numbers $n$, using the Principle of Mathematical Induction.

Proof:

We prove the statement $P(n)$ using the Principle of Mathematical Induction. The principle involves two steps:

Step 1: Base Case

We verify if the statement is true for the smallest natural number, $n=1$.

For $n=1$, we evaluate the expression $41^1 - 14^1$:

$$41^1 - 14^1 = 41 - 14 = 27$$

We check if $27$ is divisible by 27.

$$27 \div 27 = 1$$

Since 27 is divisible by 27, the statement $P(1)$ is true.

Step 2: Inductive Hypothesis

Assume that the statement $P(k)$ is true for some arbitrary positive integer $k \geq 1$.

This means that $41^k - 14^k$ is divisible by 27. Therefore, we can write:

where $m$ is some integer.

From equation (i), we can express $41^k$ as:

Step 3: Inductive Step

We need to prove that the statement $P(k+1)$ is true whenever $P(k)$ is true. That is, we need to prove that $41^{k+1} - 14^{k+1}$ is divisible by 27.

Consider the expression for $P(k+1)$:

$$41^{k+1} - 14^{k+1}$$

We can rewrite $41^{k+1}$ as $41 \cdot 41^k$. Substitute the expression for $41^k$ from the Inductive Hypothesis ($41^k = 14^k + 27m$):

$$41^{k+1} - 14^{k+1} = 41 \cdot (14^k + 27m) - 14^{k+1}$$

Distribute the 41:

$$= 41 \cdot 14^k + 41 \cdot 27m - 14^{k+1}$$

Rewrite $14^{k+1}$ as $14 \cdot 14^k$:

$$= 41 \cdot 14^k + 41 \cdot 27m - 14 \cdot 14^k$$

Group the terms involving $14^k$:

$$= (41 \cdot 14^k - 14 \cdot 14^k) + 41 \cdot 27m$$

Factor out $14^k$ from the first group:

$$= (41 - 14) \cdot 14^k + 41 \cdot 27m$$

$$= 27 \cdot 14^k + 41 \cdot 27m$$

Now, factor out 27 from the expression:

$$= 27 (14^k + 41m)$$

Since $k$ is a positive integer ($k \geq 1$), $14^k$ is an integer. Since $m$ is an integer (from the Inductive Hypothesis), the term $(14^k + 41m)$ is also an integer.

Let $p = 14^k + 41m$, where $p$ is an integer. Then we have:

$$41^{k+1} - 14^{k+1} = 27p$$

This shows that $41^{k+1} - 14^{k+1}$ is a multiple of 27, which means it is divisible by 27.

Thus, we have shown that $P(k+1)$ is true whenever $P(k)$ is true.

Conclusion:

Since $P(1)$ is true (Base Case) and $P(k+1)$ is true whenever $P(k)$ is true for $k \geq 1$ (Inductive Step), by the Principle of Mathematical Induction, the statement $P(n): 41^n - 14^n$ is divisible by 27 is true for all natural numbers $n \in \mathbb{N}$.

Given:

A statement $P(n)$ concerning an inequality involving the product of fractions.

$$P(n): \left(\frac{1}{2} \cdot \frac{3}{4} \cdot \dots \cdot \frac{2n-1}{2n}\right) < \frac{1}{\sqrt{2n+1}}$$

for all natural numbers $n \in \mathbb{N}$.

To Prove:

The statement $P(n)$ is true for all natural numbers $n$, using the Principle of Mathematical Induction.

Proof:

We prove the statement $P(n)$ using the Principle of Mathematical Induction. The principle involves two steps:

Step 1: Base Case

We verify if the statement is true for the smallest natural number, $n=1$.

For $n=1$, the left-hand side (LHS) of $P(n)$ is the first term of the product.

$$LHS = \frac{1}{2}$$

The right-hand side (RHS) of $P(n)$ for $n=1$ is $\frac{1}{\sqrt{2(1)+1}}$.

$$RHS = \frac{1}{\sqrt{2+1}} = \frac{1}{\sqrt{3}}$$

We need to check if $\frac{1}{2} < \frac{1}{\sqrt{3}}$. Squaring both positive sides (which preserves the inequality direction):

$$ \left(\frac{1}{2}\right)^2 < \left(\frac{1}{\sqrt{3}}\right)^2 $$

$$ \frac{1}{4} < \frac{1}{3} $$

This inequality is true. Since $\frac{1}{4} < \frac{1}{3}$ is true, and we squared positive numbers, the original inequality $\frac{1}{2} < \frac{1}{\sqrt{3}}$ is also true.

Thus, the statement $P(1)$ is true.

Step 2: Inductive Hypothesis

Assume that the statement $P(k)$ is true for some arbitrary positive integer $k \geq 1$. That is, we assume:

Step 3: Inductive Step

We need to prove that the statement $P(k+1)$ is true whenever $P(k)$ is true. That is, we need to prove that:

$$\left(\frac{1}{2} \cdot \frac{3}{4} \cdot \dots \cdot \frac{2k-1}{2k} \cdot \frac{2(k+1)-1}{2(k+1)}\right) < \frac{1}{\sqrt{2(k+1)+1}}$$

The $(k+1)$-th term in the product is $\frac{2k+2-1}{2k+2} = \frac{2k+1}{2k+2}$.

The inequality for $P(k+1)$ is:

$$\left(\frac{1}{2} \cdot \frac{3}{4} \cdot \dots \cdot \frac{2k-1}{2k}\right) \cdot \frac{2k+1}{2k+2} < \frac{1}{\sqrt{2k+2+1}}$$

$$\left(\frac{1}{2} \cdot \frac{3}{4} \cdot \dots \cdot \frac{2k-1}{2k}\right) \cdot \frac{2k+1}{2k+2} < \frac{1}{\sqrt{2k+3}}$$

Let $LHS_k = \frac{1}{2} \cdot \frac{3}{4} \cdot \dots \cdot \frac{2k-1}{2k}$. The Inductive Hypothesis is $LHS_k < \frac{1}{\sqrt{2k+1}}$.

The LHS of $P(k+1)$ is $LHS_{k+1} = LHS_k \cdot \frac{2k+1}{2k+2}$.

From the Inductive Hypothesis (inequality i), we multiply both sides by $\frac{2k+1}{2k+2}$. Since $k \in \mathbb{N}$, $k \geq 1$, so $\frac{2k+1}{2k+2}$ is positive. Multiplying by a positive number preserves the inequality direction.

So, we have:

$$LHS_{k+1} < \frac{2k+1}{(2k+2)\sqrt{2k+1}} = \frac{\sqrt{(2k+1)^2}}{(2k+2)\sqrt{2k+1}} = \frac{\sqrt{2k+1}}{2k+2}$$

We want to show that $LHS_{k+1} < \frac{1}{\sqrt{2k+3}}$. From the above, it is sufficient to show that $\frac{\sqrt{2k+1}}{2k+2} \leq \frac{1}{\sqrt{2k+3}}$ (the strict inequality might come later). Let's prove the strict inequality directly: $\frac{\sqrt{2k+1}}{2k+2} < \frac{1}{\sqrt{2k+3}}$.

Since both sides are positive for $k \geq 1$, we can square both sides without changing the direction of the inequality:

$$ \left(\frac{\sqrt{2k+1}}{2k+2}\right)^2 < \left(\frac{1}{\sqrt{2k+3}}\right)^2 $$

$$ \frac{2k+1}{(2k+2)^2} < \frac{1}{2k+3} $$

$$ \frac{2k+1}{4(k+1)^2} < \frac{1}{2k+3} $$

Cross-multiply (all terms are positive):

$$ (2k+1)(2k+3) < 4(k+1)^2 $$

Expand both sides:

$$ 4k^2 + 6k + 2k + 3 < 4(k^2 + 2k + 1) $$

$$ 4k^2 + 8k + 3 < 4k^2 + 8k + 4 $$

Subtract $4k^2 + 8k$ from both sides:

$$ 3 < 4 $$

This inequality is true. Since all steps were performed with positive quantities and the final inequality is true, the inequality $\frac{\sqrt{2k+1}}{2k+2} < \frac{1}{\sqrt{2k+3}}$ is true for all $k \geq 1$.

Now, combining the inequalities:

$$LHS_{k+1} < \frac{\sqrt{2k+1}}{2k+2} < \frac{1}{\sqrt{2k+3}}$$

By the transitive property of inequality, we have:

$$LHS_{k+1} < \frac{1}{\sqrt{2k+3}}$$

This is the statement $P(k+1)$.

Thus, we have shown that $P(k+1)$ is true whenever $P(k)$ is true.

Conclusion:

Since $P(1)$ is true (Base Case) and $P(k+1)$ is true whenever $P(k)$ is true for $k \geq 1$ (Inductive Step), by the Principle of Mathematical Induction, the statement $P(n): (\frac{1}{2} \cdot \frac{3}{4} \cdot \dots \cdot \frac{2n-1}{2n}) < \frac{1}{\sqrt{2n+1}}$ is true for all natural numbers $n \in \mathbb{N}$.

Given:

A statement $P(n)$ concerning the divisibility of the sum of the cubes of three consecutive natural numbers by 9.

$P(n): n^3 + (n+1)^3 + (n+2)^3$ is divisible by 9, for all natural numbers $n \in \mathbb{N}$.

To Prove:

The statement $P(n)$ is true for all natural numbers $n$, using the Principle of Mathematical Induction.

Proof:

We prove the statement $P(n)$ using the Principle of Mathematical Induction. The principle involves two steps:

Step 1: Base Case

We verify if the statement is true for the smallest natural number, $n=1$.

For $n=1$, we evaluate the expression $1^3 + (1+1)^3 + (1+2)^3$:

$$1^3 + (1+1)^3 + (1+2)^3 = 1^3 + 2^3 + 3^3$$

$$= 1 + 8 + 27$$

$$= 36$$

We check if $36$ is divisible by 9.

$$36 \div 9 = 4$$

Since 36 is divisible by 9, the statement $P(1)$ is true.

Step 2: Inductive Hypothesis

Assume that the statement $P(k)$ is true for some arbitrary positive integer $k \geq 1$.

This means that $k^3 + (k+1)^3 + (k+2)^3$ is divisible by 9. Therefore, we can write:

where $m$ is some integer.

Step 3: Inductive Step

We need to prove that the statement $P(k+1)$ is true whenever $P(k)$ is true. That is, we need to prove that $(k+1)^3 + ((k+1)+1)^3 + ((k+1)+2)^3$ is divisible by 9.

Consider the expression for $P(k+1)$:

$$E(k+1) = (k+1)^3 + (k+2)^3 + (k+3)^3$$

We can relate this expression to the expression in the Inductive Hypothesis by considering the difference $E(k+1) - E(k)$, where $E(k) = k^3 + (k+1)^3 + (k+2)^3$.

$$E(k+1) - E(k) = [(k+1)^3 + (k+2)^3 + (k+3)^3] - [k^3 + (k+1)^3 + (k+2)^3]$$

$$= (k+3)^3 - k^3$$

Expand $(k+3)^3 - k^3$ using the identity $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$ with $a = k+3$ and $b = k$:

$$= ((k+3) - k)((k+3)^2 + (k+3)k + k^2)$$

$$= (3)(k^2 + 6k + 9 + k^2 + 3k + k^2)$$

Combine like terms inside the parenthesis:

$$= 3(3k^2 + 9k + 9)$$

Factor out 3 from the parenthesis:

$$= 3 \cdot 3(k^2 + 3k + 3)$$

$$= 9(k^2 + 3k + 3)$$

So, we have the relationship:

$$E(k+1) - E(k) = 9(k^2 + 3k + 3)$$

Rearrange the equation to express $E(k+1)$:

$$E(k+1) = E(k) + 9(k^2 + 3k + 3)$$

From the Inductive Hypothesis (equation i), $E(k) = k^3 + (k+1)^3 + (k+2)^3$ is divisible by 9, so $E(k) = 9m$ for some integer $m$. Substitute this into the equation for $E(k+1)$:

$$E(k+1) = 9m + 9(k^2 + 3k + 3)$$

Factor out 9 from the expression:

$$E(k+1) = 9 (m + k^2 + 3k + 3)$$

Since $k$ is a positive integer and $m$ is an integer (from the Inductive Hypothesis), the term $(m + k^2 + 3k + 3)$ is also an integer.

Let $p = m + k^2 + 3k + 3$, where $p$ is an integer. Then we have:

$$E(k+1) = 9p$$

This shows that $E(k+1) = (k+1)^3 + (k+2)^3 + (k+3)^3$ is a multiple of 9, which means it is divisible by 9.

Thus, we have shown that $P(k+1)$ is true whenever $P(k)$ is true.

Conclusion:

Since $P(1)$ is true (Base Case) and $P(k+1)$ is true whenever $P(k)$ is true for $k \geq 1$ (Inductive Step), by the Principle of Mathematical Induction, the statement $P(n): n^3 + (n+1)^3 + (n+2)^3$ is divisible by 9 is true for all natural numbers $n \in \mathbb{N}$.

Given:

A statement $P(n)$ concerning a trigonometric identity.

$P(n): \sin(x+n\pi) = (-1)^n \sin x$, for all natural numbers $n \in \mathbb{N}$ and for a real number $x$.

To Prove:

The statement $P(n)$ is true for all natural numbers $n$, using the Principle of Mathematical Induction.

Proof:

We prove the statement $P(n)$ using the Principle of Mathematical Induction. The principle involves two steps:

Step 1: Base Case

We verify if the statement is true for the smallest natural number, $n=1$.

For $n=1$, the left-hand side (LHS) of $P(n)$ is $\sin(x+1\pi)$.

$$LHS = \sin(x+\pi)$$

Using the angle addition formula for sine, $\sin(A+B) = \sin A \cos B + \cos A \sin B$, with $A=x$ and $B=\pi$:

$$LHS = \sin x \cos \pi + \cos x \sin \pi$$

Substituting the known values $\cos \pi = -1$ and $\sin \pi = 0$:

$$LHS = \sin x (-1) + \cos x (0)$$

$$LHS = -\sin x$$

The right-hand side (RHS) of $P(n)$ for $n=1$ is $(-1)^1 \sin x$.

$$RHS = (-1)^1 \sin x = -1 \sin x = -\sin x$$

Since $LHS = RHS$, the statement $P(1)$ is true.

Step 2: Inductive Hypothesis

Assume that the statement $P(k)$ is true for some arbitrary positive integer $k \geq 1$. That is, we assume:

Step 3: Inductive Step

We need to prove that the statement $P(k+1)$ is true whenever $P(k)$ is true. That is, we need to prove that $\sin(x+(k+1)\pi) = (-1)^{k+1} \sin x$.

Consider the LHS of $P(k+1)$:

$$LHS = \sin(x+(k+1)\pi)$$

We can rewrite the argument of the sine function as $(x+k\pi) + \pi$:

$$LHS = \sin((x+k\pi) + \pi)$$

Let $y = x+k\pi$. Using the angle addition formula for sine, $\sin(y+\pi) = \sin y \cos \pi + \cos y \sin \pi$, with $y=x+k\pi$ and $B=\pi$:

$$LHS = \sin(x+k\pi) \cos \pi + \cos(x+k\pi) \sin \pi$$

Substituting the known values $\cos \pi = -1$ and $\sin \pi = 0$:

$$LHS = \sin(x+k\pi) (-1) + \cos(x+k\pi) (0)$$

$$LHS = -\sin(x+k\pi)$$

Using the Inductive Hypothesis from equation (i), substitute $(-1)^k \sin x$ for $\sin(x+k\pi)$:

$$LHS = -[(-1)^k \sin x]$$

$$LHS = (-1)^1 \cdot (-1)^k \sin x$$

Using the property of exponents $a^m \cdot a^n = a^{m+n}$:

$$LHS = (-1)^{1+k} \sin x$$

$$LHS = (-1)^{k+1} \sin x$$

This is the RHS of the statement $P(k+1)$.

Thus, we have shown that $LHS = RHS$, which means $P(k+1)$ is true whenever $P(k)$ is true.

Conclusion:

Since $P(1)$ is true (Base Case) and $P(k+1)$ is true whenever $P(k)$ is true for $k \geq 1$ (Inductive Step), by the Principle of Mathematical Induction, the statement $P(n): \sin(x+n\pi) = (-1)^n \sin x$ is true for all natural numbers $n \in \mathbb{N}$.

Given:

A square matrix $A$ such that $A^2 = I$, where $I$ is the identity matrix.

A statement $P(n)$ concerning a matrix identity.

$P(n): (A+I)^n = 2^{n-1}(A+I)$, for all positive integers $n \in \mathbb{N}$.

To Prove:

The statement $P(n)$ is true for all positive integers $n$, using the Principle of Mathematical Induction.

Proof:

We prove the statement $P(n)$ using the Principle of Mathematical Induction. The principle involves two steps:

Step 1: Base Case

We verify if the statement is true for the smallest positive integer, $n=1$.

For $n=1$, the left-hand side (LHS) of $P(n)$ is $(A+I)^1$.

$$LHS = (A+I)^1 = A+I$$

The right-hand side (RHS) of $P(n)$ for $n=1$ is $2^{1-1}(A+I)$.

$$RHS = 2^{1-1}(A+I) = 2^0(A+I) = 1 \cdot (A+I) = A+I$$

Since $LHS = RHS$, the statement $P(1)$ is true.

Step 2: Inductive Hypothesis

Assume that the statement $P(k)$ is true for some arbitrary positive integer $k \geq 1$. That is, we assume:

Step 3: Inductive Step

We need to prove that the statement $P(k+1)$ is true whenever $P(k)$ is true. That is, we need to prove that $(A+I)^{k+1} = 2^{(k+1)-1}(A+I) = 2^k(A+I)$.

Consider the LHS of $P(k+1)$:

$$LHS = (A+I)^{k+1}$$

We can rewrite $(A+I)^{k+1}$ as $(A+I)^k (A+I)$.

$$LHS = (A+I)^k (A+I)$$

Using the Inductive Hypothesis from equation (i), substitute $2^{k-1}(A+I)$ for $(A+I)^k$:

$$LHS = [2^{k-1}(A+I)] (A+I)$$

Since $2^{k-1}$ is a scalar, we can write:

$$LHS = 2^{k-1} (A+I)(A+I)$$

Expand the product of matrices $(A+I)(A+I)$: Using the distributive property for matrix multiplication and the fact that $I$ is the identity matrix ($AI = A$ and $IA = A$):

$$(A+I)(A+I) = A \cdot A + A \cdot I + I \cdot A + I \cdot I$$

$$= A^2 + A + A + I^2$$

Using the given property $A^2 = I$ and the property of the identity matrix $I^2 = I$:

$$= I + A + A + I$$

Combine the terms:

$$= 2A + 2I$$

Factor out the scalar 2:

$$= 2(A+I)$$

Substitute this back into the expression for LHS:

$$LHS = 2^{k-1} [2(A+I)]$$

Using the property of exponents $a^m \cdot a^n = a^{m+n}$:

$$LHS = 2^{k-1} \cdot 2^1 (A+I)$$

$$LHS = 2^{(k-1)+1} (A+I)$$

$$LHS = 2^k (A+I)$$

This is the RHS of the statement $P(k+1)$.

Thus, we have shown that $LHS = RHS$, which means $P(k+1)$ is true whenever $P(k)$ is true.

Conclusion:

Since $P(1)$ is true (Base Case) and $P(k+1)$ is true whenever $P(k)$ is true for $k \geq 1$ (Inductive Step), by the Principle of Mathematical Induction, the statement $P(n): (A+I)^n = 2^{n-1}(A+I)$ is true for all positive integers $n \in \mathbb{N}$.

Given:

A statement $P(n)$ concerning the divisibility of an expression by 9.

$P(n): 10^n + 3 \cdot 4^{n+2} + 5$ is divisible by 9, for all positive integers $n \in \mathbb{N}$.

To Prove:

The statement $P(n)$ is true for all positive integers $n$, using the Principle of Mathematical Induction.

Proof:

We prove the statement $P(n)$ using the Principle of Mathematical Induction. The principle involves two steps:

Step 1: Base Case

We verify if the statement is true for the smallest positive integer, $n=1$.

For $n=1$, we evaluate the expression $10^1 + 3 \cdot 4^{1+2} + 5$:

$$10^1 + 3 \cdot 4^{1+2} + 5 = 10 + 3 \cdot 4^3 + 5$$

$$= 10 + 3 \cdot 64 + 5$$

$$= 10 + 192 + 5$$

$$= 207$$

We check if $207$ is divisible by 9. The sum of the digits of 207 is $2+0+7=9$, which is divisible by 9.

$$207 \div 9 = 23$$

Since 207 is divisible by 9, the statement $P(1)$ is true.

Step 2: Inductive Hypothesis

Assume that the statement $P(k)$ is true for some arbitrary positive integer $k \geq 1$.

This means that $10^k + 3 \cdot 4^{k+2} + 5$ is divisible by 9. Therefore, we can write:

where $m$ is some integer.

Step 3: Inductive Step

We need to prove that the statement $P(k+1)$ is true whenever $P(k)$ is true. That is, we need to prove that $10^{k+1} + 3 \cdot 4^{(k+1)+2} + 5$ is divisible by 9.

Consider the expression for $P(k+1)$:

$$E(k+1) = 10^{k+1} + 3 \cdot 4^{k+3} + 5$$

Consider the expression for $P(k)$ from the inductive hypothesis:

$$E(k) = 10^k + 3 \cdot 4^{k+2} + 5$$

Let's find the difference between $E(k+1)$ and $10 \cdot E(k)$:

$$E(k+1) - 10 \cdot E(k) = (10^{k+1} + 3 \cdot 4^{k+3} + 5) - 10 \cdot (10^k + 3 \cdot 4^{k+2} + 5)$$

$$= (10 \cdot 10^k + 3 \cdot 4 \cdot 4^{k+2} + 5) - (10 \cdot 10^k + 10 \cdot 3 \cdot 4^{k+2} + 10 \cdot 5)$$

$$= 10 \cdot 10^k + 12 \cdot 4^{k+2} + 5 - 10 \cdot 10^k - 30 \cdot 4^{k+2} - 50$$

Group like terms:

$$= (10 \cdot 10^k - 10 \cdot 10^k) + (12 \cdot 4^{k+2} - 30 \cdot 4^{k+2}) + (5 - 50)$$

$$= 0 - 18 \cdot 4^{k+2} - 45$$

$$= -18 \cdot 4^{k+2} - 45$$

Factor out -9 from the expression:

$$= -9 (2 \cdot 4^{k+2} + 5)$$

So, we have the relationship:

$$E(k+1) - 10 \cdot E(k) = -9 (2 \cdot 4^{k+2} + 5)$$

Rearrange the equation to express $E(k+1)$:

$$E(k+1) = 10 \cdot E(k) - 9 (2 \cdot 4^{k+2} + 5)$$

From the Inductive Hypothesis (equation i), $E(k) = 10^k + 3 \cdot 4^{k+2} + 5$ is divisible by 9, so $E(k) = 9m$ for some integer $m$. Substitute this into the equation for $E(k+1)$:

$$E(k+1) = 10 \cdot (9m) - 9 (2 \cdot 4^{k+2} + 5)$$

$$E(k+1) = 90m - 9 (2 \cdot 4^{k+2} + 5)$$

Factor out 9 from the expression:

$$E(k+1) = 9 (10m - (2 \cdot 4^{k+2} + 5))$$

$$E(k+1) = 9 (10m - 2 \cdot 4^{k+2} - 5)$$

Since $k$ is a positive integer and $m$ is an integer (from the Inductive Hypothesis), the term $(10m - 2 \cdot 4^{k+2} - 5)$ is also an integer.

Let $p = 10m - 2 \cdot 4^{k+2} - 5$, where $p$ is an integer. Then we have:

$$E(k+1) = 9p$$

This shows that $E(k+1) = 10^{k+1} + 3 \cdot 4^{k+3} + 5$ is a multiple of 9, which means it is divisible by 9.

Thus, we have shown that $P(k+1)$ is true whenever $P(k)$ is true.

Conclusion:

Since $P(1)$ is true (Base Case) and $P(k+1)$ is true whenever $P(k)$ is true for $k \geq 1$ (Inductive Step), by the Principle of Mathematical Induction, the statement $P(n): 10^n + 3 \cdot 4^{n+2} + 5$ is divisible by 9 is true for all positive integers $n \in \mathbb{N}$.

Given:

A statement $P(n)$ concerning the divisibility of the sum of the cubes of three consecutive natural numbers by 9.

The three consecutive natural numbers starting with $n$ are $n$, $n+1$, and $n+2$. Their cubes are $n^3$, $(n+1)^3$, and $(n+2)^3$.

$P(n): n^3 + (n+1)^3 + (n+2)^3$ is divisible by 9, for all natural numbers $n \in \mathbb{N}$.

To Prove:

The statement $P(n)$ is true for all natural numbers $n$, using the Principle of Mathematical Induction.

Proof:

We prove the statement $P(n)$ using the Principle of Mathematical Induction. The principle involves two steps:

Step 1: Base Case

We verify if the statement is true for the smallest natural number, $n=1$.

For $n=1$, we evaluate the expression $1^3 + (1+1)^3 + (1+2)^3$:

$$1^3 + (1+1)^3 + (1+2)^3 = 1^3 + 2^3 + 3^3$$

$$= 1 + 8 + 27$$

$$= 36$$

We check if $36$ is divisible by 9.

$$36 \div 9 = 4$$

Since 36 is divisible by 9, the statement $P(1)$ is true.

Step 2: Inductive Hypothesis

Assume that the statement $P(k)$ is true for some arbitrary positive integer $k \geq 1$.

This means that $k^3 + (k+1)^3 + (k+2)^3$ is divisible by 9. Therefore, we can write:

where $m$ is some integer.

Step 3: Inductive Step

We need to prove that the statement $P(k+1)$ is true whenever $P(k)$ is true. That is, we need to prove that $(k+1)^3 + ((k+1)+1)^3 + ((k+1)+2)^3$ is divisible by 9.

Consider the expression for $P(k+1)$:

$$E(k+1) = (k+1)^3 + (k+2)^3 + (k+3)^3$$

We can relate this expression to the expression in the Inductive Hypothesis, $E(k) = k^3 + (k+1)^3 + (k+2)^3$. Consider the difference $E(k+1) - E(k)$:

$$E(k+1) - E(k) = [(k+1)^3 + (k+2)^3 + (k+3)^3] - [k^3 + (k+1)^3 + (k+2)^3]$$

$$= (k+3)^3 - k^3$$

Expand $(k+3)^3 - k^3$ using the identity $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$ with $a = k+3$ and $b = k$:

$$= ((k+3) - k)((k+3)^2 + (k+3)k + k^2)$$

$$= (3)(k^2 + 6k + 9 + k^2 + 3k + k^2)$$

Combine like terms inside the parenthesis:

$$= 3(3k^2 + 9k + 9)$$

Factor out 3 from the parenthesis:

$$= 3 \cdot 3(k^2 + 3k + 3)$$

$$= 9(k^2 + 3k + 3)$$

So, we have the relationship:

$$E(k+1) - E(k) = 9(k^2 + 3k + 3)$$

Rearrange the equation to express $E(k+1)$:

$$E(k+1) = E(k) + 9(k^2 + 3k + 3)$$

From the Inductive Hypothesis (equation i), $E(k) = k^3 + (k+1)^3 + (k+2)^3$ is divisible by 9, so $E(k) = 9m$ for some integer $m$. Substitute this into the equation for $E(k+1)$:

$$E(k+1) = 9m + 9(k^2 + 3k + 3)$$

Factor out 9 from the expression:

$$E(k+1) = 9 (m + k^2 + 3k + 3)$$

Since $k$ is a positive integer and $m$ is an integer (from the Inductive Hypothesis), the term $(m + k^2 + 3k + 3)$ is also an integer.

Let $p = m + k^2 + 3k + 3$, where $p$ is an integer. Then we have:

$$E(k+1) = 9p$$

This shows that $E(k+1) = (k+1)^3 + (k+2)^3 + (k+3)^3$ is a multiple of 9, which means it is divisible by 9.

Thus, we have shown that $P(k+1)$ is true whenever $P(k)$ is true.

Conclusion:

Since $P(1)$ is true (Base Case) and $P(k+1)$ is true whenever $P(k)$ is true for $k \geq 1$ (Inductive Step), by the Principle of Mathematical Induction, the statement $P(n): n^3 + (n+1)^3 + (n+2)^3$ is divisible by 9 is true for all natural numbers $n \in \mathbb{N}$.