Chapter 4 Principle Of Mathematical Induction

To Prove:

Let the given statement be P(n). That is,

P(n): $1^2 + 2^2 + 3^2 + \dots + n^2 = \frac{n(n+1)(2n+1)}{6}$ for all $n \geq 1$.

Proof:

We shall prove this statement using the Principle of Mathematical Induction. The proof consists of three steps.

Step 1: Base Case

We need to show that the statement P(n) is true for the initial value, $n = 1$.

For $n = 1$, P(1) is:

Left Hand Side (LHS) = $1^2 = 1$.

Right Hand Side (RHS) = $\frac{1(1+1)(2(1)+1)}{6} = \frac{1 \times 2 \times 3}{6} = \frac{6}{6} = 1$.

Since LHS = RHS, the statement P(1) is true.

Step 2: Inductive Hypothesis

Let us assume that the statement P(k) is true for some arbitrary positive integer $k$.

This means we assume:

Step 3: Inductive Step

Now, we need to prove that the statement P(k+1) is also true, assuming that P(k) is true.

We need to prove that:

$1^2 + 2^2 + 3^2 + \dots + k^2 + (k+1)^2 = \frac{(k+1)((k+1)+1)(2(k+1)+1)}{6}$

Which simplifies to:

$1^2 + 2^2 + 3^2 + \dots + k^2 + (k+1)^2 = \frac{(k+1)(k+2)(2k+3)}{6}$

Let's start with the LHS of the statement P(k+1):

LHS = $(1^2 + 2^2 + 3^2 + \dots + k^2) + (k+1)^2$

Using the inductive hypothesis from equation (i), we can substitute the sum of the first k squares:

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

Now, we simplify this expression by taking $(k+1)$ as a common factor:

LHS = $(k+1) \left[ \frac{k(2k+1)}{6} + (k+1) \right]$

Take the LCM inside the brackets:

LHS = $(k+1) \left[ \frac{k(2k+1) + 6(k+1)}{6} \right]$

Expand the terms in the numerator inside the brackets:

LHS = $(k+1) \left[ \frac{2k^2 + k + 6k + 6}{6} \right]$

LHS = $(k+1) \left[ \frac{2k^2 + 7k + 6}{6} \right]$

Now, we factorize the quadratic expression $2k^2 + 7k + 6$:

$2k^2 + 7k + 6 = 2k^2 + 4k + 3k + 6 = 2k(k+2) + 3(k+2) \ $$ = (k+2)(2k+3)$

Substitute this back into the expression for the LHS:

LHS = $(k+1) \frac{(k+2)(2k+3)}{6}$

LHS = $\frac{(k+1)(k+2)(2k+3)}{6}$

This is exactly the RHS of the statement P(k+1).

Thus, P(k+1) is true whenever P(k) is true.

Conclusion

By the principle of mathematical induction, the statement P(n) is true for all natural numbers $n \geq 1$.

Hence Proved.

To Prove:

The statement $P(n): 2^n > n$ is true for all positive integers $n$ (i.e., $n \ge 1$).

Proof:

We will prove the given statement using the Principle of Mathematical Induction.

Let the given statement be $P(n)$.

$P(n): 2^n > n$

Step 1: Base Case

We need to check if the statement $P(n)$ is true for $n=1$.

For $n=1$, we have $2^1 > 1$, which is $2 > 1$.

This is true.

Thus, $P(1)$ is true.

Step 2: Inductive Hypothesis

Assume that the statement $P(k)$ is true for some arbitrary positive integer $k \ge 1$.

That is, we assume:

Step 3: Inductive Step

We need to prove that the statement $P(k+1)$ is also true, assuming $P(k)$ is true.

That is, we need to show that $2^{k+1} > k+1$.

Starting with the inductive hypothesis (i), we multiply both sides by 2:

$2^{k+1} > 2k$

Now, we can write $2k$ as $k+k$.

$2^{k+1} > k + k$

Since $k$ is a positive integer, we know that $k \ge 1$.

So, we can say that $k+k \ge k+1$.

By transitivity, since $2^{k+1} > 2k$ and $2k \ge k+1$, it follows that:

$2^{k+1} > k+1$

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

Conclusion:

By the Principle of Mathematical Induction, the statement $P(n): 2^n > n$ is true for all positive integers $n$.

To Prove:

The statement $P(n): \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + \dots + \frac{1}{n(n+1)} = \frac{n}{n+1}$ for all integers $n \geq 1$.

Proof:

We will prove the given statement using the Principle of Mathematical Induction.

Let the given statement be $P(n)$.

Step 1: Base Case

We need to check if the statement $P(n)$ is true for $n=1$.

LHS of $P(1) = \frac{1}{1(1+1)} = \frac{1}{1 \cdot 2} = \frac{1}{2}$

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

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 $k \geq 1$.

That is, assume:

Step 3: Inductive Step

We need to prove that the statement $P(k+1)$ is also true.

We need to show that:

$\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{k(k+1)} + \frac{1}{(k+1)(k+2)} = \frac{k+1}{k+2}$

Consider the LHS of $P(k+1)$:

LHS $= \left(\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{k(k+1)}\right) + \frac{1}{(k+1)(k+2)}$

Using the Inductive Hypothesis (i), we substitute the sum of the first $k$ terms:

Now, we simplify by finding a common denominator, which is $(k+1)(k+2)$.

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

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

The numerator $k^2 + 2k + 1$ is a perfect square, $(k+1)^2$.

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

Since $k \geq 1$, $k+1 \neq 0$, so we can cancel one factor of $(k+1)$ from the numerator and denominator:

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

This is exactly the RHS of $P(k+1)$.

Therefore, $P(k+1)$ is true whenever $P(k)$ is true.

Conclusion:

By the Principle of Mathematical Induction, the statement $P(n)$ is true for all integers $n \geq 1$.

To Prove:

$7^n - 3^n$ is divisible by 4 for every positive integer $n$.

Proof: Using Mathematical Induction

Let $P(n)$ be the statement "$7^n - 3^n$ is divisible by 4".

Step 1: Base Case

For $n=1$, the expression is $7^1 - 3^1 = 7 - 3 = 4$.

Since 4 is divisible by 4, $P(1)$ is true.

Step 2: Induction Hypothesis

Assume that $P(k)$ is true for some positive integer $k$. That is, assume $7^k - 3^k$ is divisible by 4.

This means we can write $7^k - 3^k = 4m$ for some integer $m$.

From this, we can express $7^k$ as $7^k = 4m + 3^k$.

Step 3: Inductive Step

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

Consider the expression $7^{k+1} - 3^{k+1}$:

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

Substitute $7^k = 4m + 3^k$ from the induction hypothesis:

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

$= 28m + 7 \cdot 3^k - 3 \cdot 3^k$

$= 28m + (7-3) \cdot 3^k$

$= 28m + 4 \cdot 3^k$

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

Since $m$ is an integer and $k$ is a positive integer, $(7m + 3^k)$ is also an integer.

Let $M = 7m + 3^k$. Then $7^{k+1} - 3^{k+1} = 4M$, where $M$ is an integer.

This shows that $7^{k+1} - 3^{k+1}$ is divisible by 4. Thus, $P(k+1)$ is true.

Conclusion:

By the principle of mathematical induction, the statement is true for all positive integers $n$.

Alternate Solution (Using Algebraic Identity)

We know the algebraic identity:

$a^n - b^n = (a-b)(a^{n-1} + a^{n-2}b + \dots + ab^{n-2} + b^{n-1})$

Let $a=7$ and $b=3$.

$7^n - 3^n = (7-3)(7^{n-1} + 7^{n-2} \cdot 3 + \dots + 7 \cdot 3^{n-2} + 3^{n-1})$

$7^n - 3^n = 4(7^{n-1} + 7^{n-2} \cdot 3 + \dots + 7 \cdot 3^{n-2} + 3^{n-1})$

Since the second factor is a sum of integers, it is an integer. Therefore, $7^n - 3^n$ is a multiple of 4.

Hence Proved.

To Prove:

The inequality $(1 + x)^n \ge (1 + nx)$ is true for all natural numbers $n$, where $x > -1$. (This is known as Bernoulli's Inequality).

Proof: Using Mathematical Induction

Let $P(n)$ be the statement "$(1 + x)^n \ge (1 + nx)$".

Step 1: Base Case

For $n=1$, the inequality becomes $(1 + x)^1 \ge (1 + 1 \cdot x)$, which simplifies to $1 + x \ge 1 + x$.

This is true. So, $P(1)$ is true.

Step 2: Induction Hypothesis

Assume that $P(k)$ is true for some natural number $k$. That is, assume:

Step 3: Inductive Step

We need to prove that $P(k+1)$ is true, i.e., $(1 + x)^{k+1} \ge (1 + (k+1)x)$.

Consider the LHS of $P(k+1)$:

$(1 + x)^{k+1} = (1 + x)^k (1 + x)$

Since we are given $x > -1$, we know that $1 + x > 0$. We can multiply both sides of our induction hypothesis (i) by $(1 + x)$ without changing the inequality direction:

$(1 + x)^k (1 + x) \ge (1 + kx)(1 + x)$

$(1 + x)^{k+1} \ge 1 + x + kx + kx^2$

$(1 + x)^{k+1} \ge 1 + (k+1)x + kx^2$

Since $k$ is a natural number ($k \ge 1$) and $x$ is a real number, the term $kx^2$ is always non-negative ($kx^2 \ge 0$).

Therefore, $1 + (k+1)x + kx^2 \ge 1 + (k+1)x$.

By transitivity, we have:

$(1 + x)^{k+1} \ge 1 + (k+1)x + kx^2 \ge 1 + (k+1)x$

This implies $(1 + x)^{k+1} \ge 1 + (k+1)x$.

Thus, $P(k+1)$ is true.

Conclusion:

By the principle of mathematical induction, the inequality is true for all natural numbers $n$.

To Prove:

$2 \cdot 7^n + 3 \cdot 5^n – 5$ is divisible by 24 for all $n \in N$.

Proof: Using Mathematical Induction

Let $P(n)$ be the statement "$2 \cdot 7^n + 3 \cdot 5^n – 5$ is divisible by 24".

Step 1: Base Case

For $n=1$, the expression is $2 \cdot 7^1 + 3 \cdot 5^1 - 5 = 14 + 15 - 5 = 24$.

Since 24 is divisible by 24, $P(1)$ is true.

Step 2: Induction Hypothesis

Assume that $P(k)$ is true for some natural number $k$. That is, assume $2 \cdot 7^k + 3 \cdot 5^k - 5$ is divisible by 24.

This means we can write $2 \cdot 7^k + 3 \cdot 5^k - 5 = 24m$ for some integer $m$.

From this, we have $2 \cdot 7^k = 24m - 3 \cdot 5^k + 5$.

Step 3: Inductive Step

We need to prove that $P(k+1)$ is true, i.e., $2 \cdot 7^{k+1} + 3 \cdot 5^{k+1} - 5$ is divisible by 24.

Consider the expression for $n=k+1$:

$2 \cdot 7^{k+1} + 3 \cdot 5^{k+1} - 5 = 7 \cdot (2 \cdot 7^k) + 3 \cdot 5 \cdot 5^k - 5$

Substitute the expression for $2 \cdot 7^k$ from the induction hypothesis:

$= 7 (24m - 3 \cdot 5^k + 5) + 15 \cdot 5^k - 5$

$= 168m - 21 \cdot 5^k + 35 + 15 \cdot 5^k - 5$

$= 168m - 6 \cdot 5^k + 30$

$= 168m - 6 \cdot (5^k - 5)$

We know $168m$ is divisible by 24. We need to check if $6(5^k - 5)$ is divisible by 24.

Consider the term $5^k - 5$. We can write $5^k - 5 = 5(5^{k-1} - 1)$.

Since $5^k - 1 = (5-1)(5^{k-1} + ... + 1)$, $5^k-1$ is always divisible by 4. So $5^k-5$ is not necessarily divisible by 4. Let's try another manipulation.

Let's rewrite the expression for $P(k+1)$ differently:

$2 \cdot 7^{k+1} + 3 \cdot 5^{k+1} - 5 = 2 \cdot 7 \cdot 7^k + 3 \cdot 5 \cdot 5^k - 5$

$= 7(2 \cdot 7^k + 3 \cdot 5^k - 5) - 21 \cdot 5^k + 35 + 15 \cdot 5^k - 5$

$= 7(24m) - 6 \cdot 5^k + 30$

$= 7(24m) - 6(5^k - 5)$

Consider $5^k - 5 = 5(5^{k-1}-1)$. The term $5^{k-1}-1$ is always a multiple of 4. So $5(5^{k-1}-1)$ is a multiple of 20, not necessarily 4.

Let's check $5^k-5 \pmod{4}$. $5 \equiv 1 \pmod{4}$, so $5^k \equiv 1^k \equiv 1 \pmod{4}$. Thus, $5^k-5 \equiv 1 - 5 \equiv -4 \equiv 0 \pmod{4}$.

So, $5^k-5$ is divisible by 4. Let $5^k-5 = 4q$ for some integer $q$.

Then the expression becomes:

$7(24m) - 6(4q) = 7(24m) - 24q = 24(7m-q)$

Since $7m-q$ is an integer, the expression is divisible by 24. Thus, $P(k+1)$ is true.

Conclusion:

By the Principle of Mathematical Induction, the statement is true for all natural numbers $n$.

Note:

There appears to be a typo in the question. The inequality $1^2 + 2^2 + \dots + n^2 > \frac{n^3}{n}$ simplifies to $1^2 + 2^2 + \dots + n^2 > n^2$, which is false for $n=1$ ($1^2 > 1^2$ is false). A common related inequality is $1^2 + 2^2 + \dots + n^2 > \frac{n^3}{3}$. We will prove this corrected statement.

To Prove:

$1^2 + 2^2 + \dots + n^2 > \frac{n^3}{3}$ for all $n \in N$.

Proof: Using Mathematical Induction

Let $P(n)$ be the statement "$1^2 + 2^2 + \dots + n^2 > \frac{n^3}{3}$".

Step 1: Base Case

For $n=1$, LHS = $1^2 = 1$. RHS = $\frac{1^3}{3} = \frac{1}{3}$.

Since $1 > \frac{1}{3}$, the statement $P(1)$ is true.

Step 2: Induction Hypothesis

Assume that $P(k)$ is true for some natural number $k$. That is, assume:

Step 3: Inductive Step

We need to prove that $P(k+1)$ is true, i.e., $1^2 + 2^2 + \dots + (k+1)^2 > \frac{(k+1)^3}{3}$.

Consider the LHS of $P(k+1)$:

LHS $= (1^2 + 2^2 + \dots + k^2) + (k+1)^2$

Using the induction hypothesis (i), we can say:

LHS $> \frac{k^3}{3} + (k+1)^2$

Now, we need to show that $\frac{k^3}{3} + (k+1)^2 > \frac{(k+1)^3}{3}$.

Consider the difference: $\left(\frac{k^3}{3} + (k+1)^2\right) - \frac{(k+1)^3}{3}$

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

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

$= \frac{k^3 + 3k^2+6k+3 - k^3-3k^2-3k-1}{3}$

$= \frac{3k+2}{3}$

Since $k$ is a natural number ($k \ge 1$), $3k+2$ is positive. Therefore, $\frac{3k+2}{3} > 0$.

This shows that $\frac{k^3}{3} + (k+1)^2 > \frac{(k+1)^3}{3}$.

By transitivity, since LHS $> \frac{k^3}{3} + (k+1)^2$ and $\frac{k^3}{3} + (k+1)^2 > \frac{(k+1)^3}{3}$, we have:

LHS $> \frac{(k+1)^3}{3}$

Thus, $P(k+1)$ is true.

Conclusion:

By the Principle of Mathematical Induction, the statement is true for all natural numbers $n$.

To Prove:

The rule of exponents $(ab)^n = a^n b^n$ is true for every natural number $n$.

Proof: Using Mathematical Induction

Let $P(n)$ be the statement "$(ab)^n = a^n b^n$".

Step 1: Base Case

For $n=1$, the statement is $(ab)^1 = a^1 b^1$, which simplifies to $ab = ab$.

This is true. So, $P(1)$ is true.

Step 2: Induction Hypothesis

Assume that $P(k)$ is true for some natural number $k$. That is, assume:

Step 3: Inductive Step

We need to prove that $P(k+1)$ is true, i.e., $(ab)^{k+1} = a^{k+1} b^{k+1}$.

Consider the LHS of $P(k+1)$:

LHS $= (ab)^{k+1}$

Using the definition of exponents, we can write this as:

LHS $= (ab)^k \cdot (ab)$

Now, apply the induction hypothesis (i) to the term $(ab)^k$:

LHS $= (a^k b^k) \cdot (ab)$

Using the associative and commutative properties of multiplication, we can rearrange the terms:

LHS $= (a^k \cdot a) \cdot (b^k \cdot b)$

Using the definition of exponents again:

LHS $= a^{k+1} \cdot b^{k+1}$

This is the RHS of the statement $P(k+1)$.

Thus, $P(k+1)$ is true whenever $P(k)$ is true.

Conclusion:

By the Principle of Mathematical Induction, the rule of exponents $(ab)^n = a^n b^n$ is true for all natural numbers $n$.

Given:

The statement $1 + 3 + 3^2 + \dots + 3^{n - 1} = \frac{(3^n - 1)}{2}$.

$n$ is a natural number ($n \in \{1, 2, 3, \dots \}$).

To Prove:

$1 + 3 + 3^2 + \dots + 3^{n - 1} = \frac{(3^n - 1)}{2}$ for all natural numbers $n$, using the principle of mathematical induction.

Proof: Using Mathematical Induction

Let $P(n)$ be the statement "$1 + 3 + 3^2 + \dots + 3^{n - 1} = \frac{(3^n - 1)}{2}$".

Base Case (n=1):

For $n=1$, the left side of the statement is the sum up to the term $3^{1-1} = 3^0 = 1$.

LHS $= 1$

The right side of the statement for $n=1$ is:

RHS $= \frac{(3^1 - 1)}{2} = \frac{3 - 1}{2} = \frac{2}{2} = 1$

Comparing LHS and RHS:

$1 = 1$

This is true. So, $P(1)$ is true.

Induction Hypothesis:

Assume that $P(k)$ is true for some natural number $k$. That is, assume:

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

This is our induction hypothesis.

Inductive Step (n=k+1):

We need to prove that $P(k+1)$ is true, i.e., we need to show that $1 + 3 + 3^2 + \dots + 3^{(k+1) - 1} = \frac{(3^{k+1} - 1)}{2}$.

The term $3^{(k+1) - 1}$ is $3^k$. The statement for $n=k+1$ is $1 + 3 + 3^2 + \dots + 3^{k - 1} + 3^k = \frac{(3^{k+1} - 1)}{2}$.

Consider the left side of the statement for $n=k+1$:

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

Using the induction hypothesis, we can replace the sum in the parenthesis with $\frac{(3^k - 1)}{2}$:

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

Now, we simplify the expression to match the RHS for $n=k+1$:

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

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

LHS $= \frac{(1 + 2) \cdot 3^k - 1}{2}$

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

Using the rule of exponents $a^m \cdot a^n = a^{m+n}$:

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

This is the right side of the statement for $n=k+1$ (RHS $= \frac{(3^{k+1} - 1)}{2}$).

Since LHS = RHS, the statement $P(k+1)$ is true.

Conclusion:

By the Principle of Mathematical Induction, the statement $P(n): 1 + 3 + 3^2 + \dots + 3^{n - 1} = \frac{(3^n - 1)}{2}$ is true for all natural numbers $n$.

Given:

The statement $1^3 + 2^3 + 3^3 + \dots + n^3 = \left( \frac{n(n+1)}{2} \right)^2$.

$n$ is a natural number ($n \in \{1, 2, 3, \dots \}$).

To Prove:

$1^3 + 2^3 + 3^3 + \dots + n^3 = \left( \frac{n(n+1)}{2} \right)^2$ for all natural numbers $n$, using the principle of mathematical induction.

Proof: Using Mathematical Induction

Let $P(n)$ be the statement "$1^3 + 2^3 + \dots + n^3 = \left( \frac{n(n+1)}{2} \right)^2$".

Base Case (n=1):

For $n=1$, the left side of the statement is:

LHS $= 1^3 = 1$

The right side of the statement for $n=1$ is:

RHS $= \left( \frac{1(1+1)}{2} \right)^2 = \left( \frac{1(2)}{2} \right)^2 = \left( \frac{2}{2} \right)^2 = 1^2 = 1$

Comparing LHS and RHS:

$1 = 1$

This is true. So, $P(1)$ is true.

Induction Hypothesis:

Assume that $P(k)$ is true for some natural number $k$. That is, assume:

$1^3 + 2^3 + 3^3 + \dots + k^3 = \left( \frac{k(k+1)}{2} \right)^2$

This is our induction hypothesis.

Inductive Step (n=k+1):

We need to prove that $P(k+1)$ is true, i.e., we need to show that $1^3 + 2^3 + 3^3 + \dots + (k+1)^3 = \left( \frac{(k+1)((k+1)+1)}{2} \right)^2 = \left( \frac{(k+1)(k+2)}{2} \right)^2$.

Consider the left side of the statement for $n=k+1$:

LHS $= (1^3 + 2^3 + 3^3 + \dots + k^3) + (k+1)^3$

Using the induction hypothesis, we can replace the sum in the parenthesis with $\left( \frac{k(k+1)}{2} \right)^2$:

LHS $= \left( \frac{k(k+1)}{2} \right)^2 + (k+1)^3$

Now, we simplify the expression:

LHS $= \frac{k^2(k+1)^2}{4} + (k+1)^3$

To combine the terms, find a common denominator (4):

LHS $= \frac{k^2(k+1)^2}{4} + \frac{4(k+1)^3}{4}$

Factor out the common term $(k+1)^2$ from the numerator:

LHS $= \frac{(k+1)^2 [k^2 + 4(k+1)]}{4}$

Expand the term inside the square brackets:

LHS $= \frac{(k+1)^2 [k^2 + 4k + 4]}{4}$

Recognize that $k^2 + 4k + 4$ is a perfect square trinomial, specifically $(k+2)^2$:

LHS $= \frac{(k+1)^2 (k+2)^2}{4}$

We can rewrite the denominator as $2^2$:

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

Using the rule of exponents $\frac{a^m b^m}{c^m} = \left( \frac{ab}{c} \right)^m$:

LHS $= \left( \frac{(k+1)(k+2)}{2} \right)^2$

This is the right side of the statement for $n=k+1$ (RHS $= \left( \frac{(k+1)(k+2)}{2} \right)^2$).

Since LHS = RHS, the statement $P(k+1)$ is true.

Conclusion:

By the Principle of Mathematical Induction, the statement $P(n): 1^3 + 2^3 + 3^3 + \dots + n^3 = \left( \frac{n(n+1)}{2} \right)^2$ is true for all natural numbers $n$.

Given:

The statement $1 + \frac{1}{(1 + 2)} + \frac{1}{(1 + 2 + 3)} + \dots + \frac{1}{(1 + 2 + 3 + \dots + n)} = \frac{2n}{(n + 1)}$.

$n$ is a natural number ($n \in \{1, 2, 3, \dots \}$).

To Prove:

$1 + \frac{1}{(1 + 2)} + \frac{1}{(1 + 2 + 3)} + \dots + \frac{1}{(1 + 2 + 3 + \dots + n)} = \frac{2n}{(n + 1)}$ for all natural numbers $n$, using the principle of mathematical induction.

Proof: Using Mathematical Induction

Let $P(n)$ be the statement "$1 + \frac{1}{(1 + 2)} + \frac{1}{(1 + 2 + 3)} + \dots + \frac{1}{(1 + 2 + 3 + \dots + n)} = \frac{2n}{(n + 1)}$".

The sum of the first $r$ natural numbers is $1 + 2 + \dots + r = \frac{r(r+1)}{2}$. So, the $r$-th term in the sum is $\frac{1}{\frac{r(r+1)}{2}} = \frac{2}{r(r+1)}$.

Thus, $P(n)$ can be written as: $\sum\limits_{r=1}^{n} \frac{2}{r(r+1)} = \frac{2n}{n+1}$.

Base Case (n=1):

For $n=1$, the left side of the statement is the first term, $r=1$.

LHS $= 1$.

The right side of the statement for $n=1$ is:

RHS $= \frac{2 \cdot 1}{(1 + 1)} = \frac{2}{2} = 1$.

Comparing LHS and RHS:

$1 = 1$

This is true. So, $P(1)$ is true.

Induction Hypothesis:

Assume that $P(k)$ is true for some natural number $k$. That is, assume:

$1 + \frac{1}{(1 + 2)} + \frac{1}{(1 + 2 + 3)} + \dots + \frac{1}{(1 + 2 + 3 + \dots + k)} = \frac{2k}{(k + 1)}$

This can be written as $\sum\limits_{r=1}^{k} \frac{2}{r(r+1)} = \frac{2k}{k+1}$.

Inductive Step (n=k+1):

We need to prove that $P(k+1)$ is true, i.e., we need to show that:

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

Consider the left side of the statement for $n=k+1$:

LHS $= \left( 1 + \frac{1}{(1 + 2)} + \dots + \frac{1}{(1 + 2 + \dots + k)} \right) + \frac{1}{(1 + 2 + \dots + (k+1))}$

Using the induction hypothesis, replace the sum in the parenthesis:

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

The denominator of the last term is the sum of the first $(k+1)$ natural numbers:

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

So, the last term is $\frac{1}{\frac{(k+1)(k+2)}{2}} = \frac{2}{(k+1)(k+2)}$.

Substitute this into the expression for LHS:

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

Combine the fractions by finding a common denominator $(k+1)(k+2)$:

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

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

Expand the numerator:

Numerator $= 2k^2 + 4k + 2$

Factor out 2 from the numerator:

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

Recognize the perfect square trinomial $k^2 + 2k + 1 = (k+1)^2$:

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

Substitute the factored numerator back into the LHS expression:

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

Cancel out one factor of $(k+1)$ from the numerator and denominator (since $k \in \mathbb{N}$, $k+1 \ne 0$):

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

This is the right side of the statement for $n=k+1$ (RHS $= \frac{2(k+1)}{(k+2)}$).

Since LHS = RHS, the statement $P(k+1)$ is true.

Conclusion:

By the Principle of Mathematical Induction, the statement $P(n): 1 + \frac{1}{(1 + 2)} + \frac{1}{(1 + 2 + 3)} + \dots + \frac{1}{(1 + 2 + 3 + \dots + n)} = \frac{2n}{(n + 1)}$ is true for all natural numbers $n$.

Given:

The statement $1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + \dots + n(n + 1)(n + 2) = \frac{n(n + 1)(n + 2)(n + 3)}{4}$.

$n$ is a natural number ($n \in \{1, 2, 3, \dots \}$).

To Prove:

$1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + \dots + n(n + 1)(n + 2) = \frac{n(n + 1)(n + 2)(n + 3)}{4}$ for all natural numbers $n$, using the principle of mathematical induction.

Proof: Using Mathematical Induction

Let $P(n)$ be the statement "$1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + \dots + n(n + 1)(n + 2) = \frac{n(n + 1)(n + 2)(n + 3)}{4}$".

Base Case (n=1):

For $n=1$, the left side of the statement is the first term:

LHS $= 1 \cdot (1+1) \cdot (1+2) = 1 \cdot 2 \cdot 3 = 6$

The right side of the statement for $n=1$ is:

RHS $= \frac{1(1+1)(1+2)(1+3)}{4} = \frac{1 \cdot 2 \cdot 3 \cdot 4}{4} = \frac{24}{4} = 6$

Comparing LHS and RHS:

$6 = 6$

This is true. So, $P(1)$ is true.

Induction Hypothesis:

Assume that $P(k)$ is true for some natural number $k$. That is, assume:

$1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + \dots + k(k+1)(k+2) = \frac{k(k+1)(k+2)(k+3)}{4}$

This is our induction hypothesis.

Inductive Step (n=k+1):

We need to prove that $P(k+1)$ is true, i.e., we need to show that:

$1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + \dots + (k+1)((k+1)+1)((k+1)+2) \ $$ = \frac{(k+1)((k+1)+1)((k+1)+2)((k+1)+3)}{4}$

This simplifies to showing:

$1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + \dots + (k+1)(k+2)(k+3) = \frac{(k+1)(k+2)(k+3)(k+4)}{4}$.

Consider the left side of the statement for $n=k+1$:

LHS $= (1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + \dots + k(k+1)(k+2)) + (k+1)(k+2) \ $$ (k+3)$

Using the induction hypothesis, we can replace the sum in the parenthesis with $\frac{k(k+1)(k+2)(k+3)}{4}$:

LHS $= \frac{k(k+1)(k+2)(k+3)}{4} + (k+1) \ $$ (k+2)(k+3)$

Now, we simplify the expression. Factor out the common term $(k+1)(k+2)(k+3)$:

LHS $= (k+1)(k+2)(k+3) \left( \frac{k}{4} + 1 \right)$

Combine the terms inside the parenthesis:

LHS $= (k+1)(k+2)(k+3) \left( \frac{k}{4} + \frac{4}{4} \right)$

LHS $= (k+1)(k+2)(k+3) \left( \frac{k+4}{4} \right)$

Rewrite the expression:

LHS $= \frac{(k+1)(k+2)(k+3)(k+4)}{4}$

This is the right side of the statement for $n=k+1$ (RHS $= \frac{(k+1)(k+2)(k+3)(k+4)}{4}$).

Since LHS = RHS, the statement $P(k+1)$ is true.

Conclusion:

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 $n$.

Given:

The statement $1 \cdot 3 + 2 \cdot 3^2 + 3 \cdot 3^3 + \dots + n \cdot 3^n = \frac{(2n - 1)3^{n+1} + 3}{4}$.

$n$ is a natural number ($n \in \{1, 2, 3, \dots \}$).

To Prove:

$1 \cdot 3 + 2 \cdot 3^2 + 3 \cdot 3^3 + \dots + n \cdot 3^n = \frac{(2n - 1)3^{n+1} + 3}{4}$ for all natural numbers $n$, using the principle of mathematical induction.

Proof: Using Mathematical Induction

Let $P(n)$ be the statement "$1 \cdot 3 + 2 \cdot 3^2 + 3 \cdot 3^3 + \dots + n \cdot 3^n = \frac{(2n - 1)3^{n+1} + 3}{4}$".

Base Case (n=1):

For $n=1$, the left side of the statement is the first term:

LHS $= 1 \cdot 3^1 = 3$

The right side of the statement for $n=1$ is:

RHS $= \frac{(2 \cdot 1 - 1)3^{1+1} + 3}{4} = \frac{(2 - 1)3^2 + 3}{4} = \frac{1 \cdot 9 + 3}{4} = \frac{9 + 3}{4} = \frac{12}{4} = 3$

Comparing LHS and RHS:

$3 = 3$

This is true. So, $P(1)$ is true.

Induction Hypothesis:

Assume that $P(k)$ is true for some natural number $k$. That is, assume:

$1 \cdot 3 + 2 \cdot 3^2 + 3 \cdot 3^3 + \dots + k \cdot 3^k = \frac{(2k - 1)3^{k+1} + 3}{4}$

This is our induction hypothesis.

Inductive Step (n=k+1):

We need to prove that $P(k+1)$ is true, i.e., we need to show that:

$1 \cdot 3 + 2 \cdot 3^2 + \dots + k \cdot 3^k + (k+1) \cdot 3^{k+1} = \frac{(2(k+1) - 1)3^{(k+1)+1} + 3}{4}$

This simplifies to showing:

$1 \cdot 3 + 2 \cdot 3^2 + \dots + k \cdot 3^k + (k+1) \cdot 3^{k+1} = \frac{(2k + 1)3^{k+2} + 3}{4}$.

Consider the left side of the statement for $n=k+1$:

LHS $= (1 \cdot 3 + 2 \cdot 3^2 + \dots + k \cdot 3^k) + (k+1) \cdot 3^{k+1}$

Using the induction hypothesis, we can replace the sum in the parenthesis with $\frac{(2k - 1)3^{k+1} + 3}{4}$:

LHS $= \frac{(2k - 1)3^{k+1} + 3}{4} + (k+1) \cdot 3^{k+1}$

Now, we simplify the expression:

LHS $= \frac{(2k - 1)3^{k+1} + 3 + 4(k+1) \cdot 3^{k+1}}{4}$

LHS $= \frac{(2k - 1)3^{k+1} + 4(k+1)3^{k+1} + 3}{4}$

Factor out $3^{k+1}$ from the terms containing it in the numerator:

LHS $= \frac{3^{k+1} [(2k - 1) + 4(k+1)] + 3}{4}$

Simplify the expression inside the square brackets:

$ (2k - 1) + 4(k+1) = 2k - 1 + 4k + 4 = 6k + 3 $

Substitute this back into the LHS expression:

LHS $= \frac{3^{k+1} (6k + 3) + 3}{4}$

Factor out 3 from $(6k + 3)$:

LHS $= \frac{3^{k+1} \cdot 3(2k + 1) + 3}{4}$

Using the rule of exponents $a^m \cdot a^n = a^{m+n}$ ($3^{k+1} \cdot 3 = 3^{k+1+1} = 3^{k+2}$):

LHS $= \frac{3^{k+2} (2k + 1) + 3}{4}$

Rearranging the terms in the numerator:

LHS $= \frac{(2k + 1)3^{k+2} + 3}{4}$

This is the right side of the statement for $n=k+1$ (RHS $= \frac{(2k + 1)3^{k+2} + 3}{4}$).

Since LHS = RHS, the statement $P(k+1)$ is true.

Conclusion:

By the Principle of Mathematical Induction, the statement $P(n): 1 \cdot 3 + 2 \cdot 3^2 + 3 \cdot 3^3 + \dots + n \cdot 3^n = \frac{(2n - 1)3^{n+1} + 3}{4}$ is true for all natural numbers $n$.

Given:

The statement $1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \dots + n(n+1) = \frac{n(n+1)(n+2)}{3}$.

$n$ is a natural number ($n \in \{1, 2, 3, \dots \}$).

To Prove:

$1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \dots + n(n+1) = \frac{n(n+1)(n+2)}{3}$ for all natural numbers $n$, using the principle of mathematical induction.

Proof: Using Mathematical Induction

Let $P(n)$ be the statement "$1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \dots + n(n+1) = \frac{n(n+1)(n+2)}{3}$".

Base Case (n=1):

For $n=1$, the left side of the statement is the first term:

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

The right side of the statement for $n=1$ is:

RHS $= \frac{1(1+1)(1+2)}{3} = \frac{1 \cdot 2 \cdot 3}{3} = \frac{6}{3} = 2$

Comparing LHS and RHS:

$2 = 2$

This is true. So, $P(1)$ is true.

Induction Hypothesis:

Assume that $P(k)$ is true for some natural number $k$. That is, assume:

$1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \dots + k(k+1) = \frac{k(k+1)(k+2)}{3}$

This is our induction hypothesis.

Inductive Step (n=k+1):

We need to prove that $P(k+1)$ is true, i.e., we need to show that:

$1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \dots + (k+1)((k+1)+1) \ $$ = \frac{(k+1)((k+1)+1)((k+1)+2)}{3}$

This simplifies to showing:

$1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \dots + (k+1)(k+2) = \frac{(k+1)(k+2)(k+3)}{3}$.

Consider the left side of the statement for $n=k+1$:

LHS $= (1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \dots + k(k+1)) + (k+1)(k+2)$

Using the induction hypothesis, we can replace the sum in the parenthesis with $\frac{k(k+1)(k+2)}{3}$:

LHS $= \frac{k(k+1)(k+2)}{3} + (k+1)(k+2)$

Now, we simplify the expression. Factor out the common term $(k+1)(k+2)$:

LHS $= (k+1)(k+2) \left( \frac{k}{3} + 1 \right)$

Combine the terms inside the parenthesis:

LHS $= (k+1)(k+2) \left( \frac{k}{3} + \frac{3}{3} \right)$

LHS $= (k+1)(k+2) \left( \frac{k+3}{3} \right)$

Rewrite the expression:

LHS $= \frac{(k+1)(k+2)(k+3)}{3}$

This is the right side of the statement for $n=k+1$ (RHS $= \frac{(k+1)(k+2)(k+3)}{3}$).

Since LHS = RHS, the statement $P(k+1)$ is true.

Conclusion:

By the Principle of Mathematical Induction, the statement $P(n): 1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \dots + n(n+1) = \frac{n(n+1)(n+2)}{3}$ is true for all natural numbers $n$.

Given:

The statement $1 \cdot 3 + 3 \cdot 5 + 5 \cdot 7 + \dots + (2n - 1)(2n + 1) = \frac{n(4n^2 + 6n - 1)}{3}$.

$n$ is a natural number ($n \in \{1, 2, 3, \dots \}$).

To Prove:

$1 \cdot 3 + 3 \cdot 5 + 5 \cdot 7 + \dots + (2n - 1)(2n + 1) = \frac{n(4n^2 + 6n - 1)}{3}$ for all natural numbers $n$, using the principle of mathematical induction.

Proof: Using Mathematical Induction

Let $P(n)$ be the statement "$1 \cdot 3 + 3 \cdot 5 + 5 \cdot 7 + \dots + (2n - 1)(2n + 1) = \frac{n(4n^2 + 6n - 1)}{3}$".

Base Case (n=1):

For $n=1$, the left side of the statement is the first term:

LHS $= (2 \cdot 1 - 1)(2 \cdot 1 + 1) = (2 - 1)(2 + 1) = 1 \cdot 3 = 3$

The right side of the statement for $n=1$ is:

RHS $= \frac{1(4(1)^2 + 6(1) - 1)}{3} = \frac{1(4 + 6 - 1)}{3} = \frac{1(9)}{3} = \frac{9}{3} = 3$

Comparing LHS and RHS:

$3 = 3$

This is true. So, $P(1)$ is true.

Induction Hypothesis:

Assume that $P(k)$ is true for some natural number $k$. That is, assume:

$1 \cdot 3 + 3 \cdot 5 + 5 \cdot 7 + \dots + (2k - 1)(2k + 1) = \frac{k(4k^2 + 6k - 1)}{3}$

This is our induction hypothesis.

Inductive Step (n=k+1):

We need to prove that $P(k+1)$ is true, i.e., we need to show that:

$1 \cdot 3 + 3 \cdot 5 + \dots + (2k - 1)(2k + 1) + (2(k+1) - 1)(2(k+1) + 1) \ $$ = \frac{(k+1)(4(k+1)^2 + 6(k+1) - 1)}{3}$.

The $(k+1)$-th term is $(2(k+1) - 1)(2(k+1) + 1) = (2k + 2 - 1) \ $$ (2k + 2 + 1) = (2k + 1)(2k + 3)$.

Consider the left side of the statement for $n=k+1$:

LHS $= (1 \cdot 3 + 3 \cdot 5 + \dots + (2k - 1)(2k + 1)) + (2k + 1)(2k + 3)$

Using the induction hypothesis, we can replace the sum in the parenthesis with $\frac{k(4k^2 + 6k - 1)}{3}$:

LHS $= \frac{k(4k^2 + 6k - 1)}{3} + (2k + 1)(2k + 3)$

Now, we simplify the expression:

LHS $= \frac{k(4k^2 + 6k - 1) + 3(2k + 1)(2k + 3)}{3}$

Expand the terms in the numerator:

$k(4k^2 + 6k - 1) = 4k^3 + 6k^2 - k$

$3(2k + 1)(2k + 3) = 3(4k^2 + 6k + 2k + 3) = 3(4k^2 + 8k + 3) \ $$ = 12k^2 + 24k + 9$

Add the expanded terms in the numerator:

Numerator $= (4k^3 + 6k^2 - k) + (12k^2 + 24k + 9) \ $$ = 4k^3 + 18k^2 + 23k + 9$

So, LHS $= \frac{4k^3 + 18k^2 + 23k + 9}{3}$.

Now, consider the right side of the statement for $n=k+1$:

RHS $= \frac{(k+1)(4(k+1)^2 + 6(k+1) - 1)}{3}$

Expand the terms inside the parenthesis:

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

$4(k+1)^2 = 4(k^2 + 2k + 1) = 4k^2 + 8k + 4$

$6(k+1) = 6k + 6$

RHS $= \frac{(k+1)(4k^2 + 8k + 4 + 6k + 6 - 1)}{3}$

RHS $= \frac{(k+1)(4k^2 + 14k + 9)}{3}$

Expand the numerator:

$(k+1)(4k^2 + 14k + 9) = k(4k^2 + 14k + 9) + 1(4k^2 + 14k + 9)$

$= 4k^3 + 14k^2 + 9k + 4k^2 + 14k + 9$

$= 4k^3 + 18k^2 + 23k + 9$

So, RHS $= \frac{4k^3 + 18k^2 + 23k + 9}{3}$.

Comparing the simplified LHS and RHS:

LHS $= \frac{4k^3 + 18k^2 + 23k + 9}{3}$

RHS $= \frac{4k^3 + 18k^2 + 23k + 9}{3}$

Since LHS = RHS, the statement $P(k+1)$ is true.

Conclusion:

By the Principle of Mathematical Induction, the statement $P(n): 1 \cdot 3 + 3 \cdot 5 + 5 \cdot 7 + \dots + (2n - 1)(2n + 1) = \frac{n(4n^2 + 6n - 1)}{3}$ is true for all natural numbers $n$.

Given:

The statement $1 \cdot 2 + 2 \cdot 2^2 + 3 \cdot 2^3 + \dots + n \cdot 2^n = (n - 1) 2^{n + 1} + 2$.

$n$ is a natural number ($n \in \{1, 2, 3, \dots \}$).

To Prove:

$1 \cdot 2 + 2 \cdot 2^2 + 3 \cdot 2^3 + \dots + n \cdot 2^n = (n - 1) 2^{n + 1} + 2$ for all natural numbers $n$, using the principle of mathematical induction.

Proof: Using Mathematical Induction

Let $P(n)$ be the statement "$1 \cdot 2 + 2 \cdot 2^2 + 3 \cdot 2^3 + \dots + n \cdot 2^n = (n - 1) 2^{n + 1} + 2$".

Base Case (n=1):

For $n=1$, the left side of the statement is the first term:

LHS $= 1 \cdot 2^1 = 2$

The right side of the statement for $n=1$ is:

RHS $= (1 - 1) 2^{1 + 1} + 2 = 0 \cdot 2^2 + 2 = 0 \cdot 4 + 2 = 0 + 2 = 2$

Comparing LHS and RHS:

$2 = 2$

This is true. So, $P(1)$ is true.

Induction Hypothesis:

Assume that $P(k)$ is true for some natural number $k$. That is, assume:

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

This is our induction hypothesis.

Inductive Step (n=k+1):

We need to prove that $P(k+1)$ is true, i.e., we need to show that:

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

This simplifies to showing:

$1 \cdot 2 + 2 \cdot 2^2 + 3 \cdot 2^3 + \dots + (k+1) \cdot 2^{k+1} = k \cdot 2^{k + 2} + 2$.

Consider the left side of the statement for $n=k+1$:

LHS $= (1 \cdot 2 + 2 \cdot 2^2 + 3 \cdot 2^3 + \dots + k \cdot 2^k) + (k+1) \cdot 2^{k+1}$

Using the induction hypothesis, we can replace the sum in the parenthesis with $(k - 1) 2^{k + 1} + 2$:

LHS $= ((k - 1) 2^{k + 1} + 2) + (k+1) \cdot 2^{k+1}$

Group the terms containing $2^{k+1}$:

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

Factor out the common term $2^{k+1}$:

LHS $= 2^{k + 1} [(k - 1) + (k+1)] + 2$

Simplify the expression inside the square brackets:

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

Substitute this back into the expression for LHS:

LHS $= 2^{k + 1} (2k) + 2$

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

Using the rule of exponents $a^m \cdot a^n = a^{m+n}$, $2 \cdot 2^{k+1} = 2^1 \cdot 2^{k+1} = 2^{1+(k+1)} = 2^{k+2}$.

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

LHS $= k \cdot 2^{k + 2} + 2$

This is the right side of the statement for $n=k+1$ (RHS $= k \cdot 2^{k + 2} + 2$).

Since LHS = RHS, the statement $P(k+1)$ is true.

Conclusion:

By the Principle of Mathematical Induction, the statement $P(n): 1 \cdot 2 + 2 \cdot 2^2 + 3 \cdot 2^3 + \dots + n \cdot 2^n = (n - 1) 2^{n + 1} + 2$ is true for all natural numbers $n$.

Given:

The statement $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots + \frac{1}{2^{n}} = 1 - \frac{1}{2^{n}}$.

$n$ is a natural number ($n \in \{1, 2, 3, \dots \}$).

To Prove:

$\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots + \frac{1}{2^{n}} = 1 - \frac{1}{2^{n}}$ for all natural numbers $n$, using the principle of mathematical induction.

Proof: Using Mathematical Induction

Let $P(n)$ be the statement "$\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots + \frac{1}{2^{n}} = 1 - \frac{1}{2^{n}}$".

Base Case (n=1):

For $n=1$, the left side of the statement is the first term:

LHS $= \frac{1}{2^1} = \frac{1}{2}$

The right side of the statement for $n=1$ is:

RHS $= 1 - \frac{1}{2^1} = 1 - \frac{1}{2} = \frac{1}{2}$

Comparing LHS and RHS:

$\frac{1}{2} = \frac{1}{2}$

This is true. So, $P(1)$ is true.

Induction Hypothesis:

Assume that $P(k)$ is true for some natural number $k$. That is, assume:

$\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots + \frac{1}{2^{k}} = 1 - \frac{1}{2^{k}}$

Inductive Step (n=k+1):

We need to prove that $P(k+1)$ is true, i.e., we need to show that:

$\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots + \frac{1}{2^{k}} + \frac{1}{2^{k+1}} = 1 - \frac{1}{2^{k+1}}$

Consider the left side of the statement for $n=k+1$:

LHS $= \left( \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots + \frac{1}{2^{k}} \right) + \frac{1}{2^{k+1}}$

Using the induction hypothesis, we can replace the sum in the parenthesis with $1 - \frac{1}{2^{k}}$:

LHS $= \left( 1 - \frac{1}{2^{k}} \right) + \frac{1}{2^{k+1}}$

Now, we simplify the expression:

LHS $= 1 - \frac{1}{2^{k}} + \frac{1}{2^{k+1}}$

To combine the fractions, rewrite $\frac{1}{2^k}$ with denominator $2^{k+1}$:

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

Substitute this back into the expression for LHS:

LHS $= 1 - \frac{2}{2^{k+1}} + \frac{1}{2^{k+1}}$

Combine the fractions:

LHS $= 1 + \frac{-2 + 1}{2^{k+1}}$

LHS $= 1 + \frac{-1}{2^{k+1}}$

LHS $= 1 - \frac{1}{2^{k+1}}$

This is the right side of the statement for $n=k+1$ (RHS $= 1 - \frac{1}{2^{k+1}}$).

Since LHS = RHS, the statement $P(k+1)$ is true.

Conclusion:

By the Principle of Mathematical Induction, the statement $P(n): \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots + \frac{1}{2^{n}} = 1 - \frac{1}{2^{n}}$ is true for all natural numbers $n$.

Given:

The statement $\frac{1}{2 \cdot 5} + \frac{1}{5 \cdot 8} + \frac{1}{8 \cdot 11} + \dots + \frac{1}{(3n - 1)(3n + 2)} = \frac{n}{6n + 4}$.

$n$ is a natural number ($n \in \{1, 2, 3, \dots \}$).

To Prove:

$\frac{1}{2 \cdot 5} + \frac{1}{5 \cdot 8} + \frac{1}{8 \cdot 11} + \dots + \frac{1}{(3n - 1)(3n + 2)} = \frac{n}{6n + 4}$ for all natural numbers $n$, using the principle of mathematical induction.

Proof: Using Mathematical Induction

Let $P(n)$ be the statement "$\frac{1}{2 \cdot 5} + \frac{1}{5 \cdot 8} + \frac{1}{8 \cdot 11} + \dots + \frac{1}{(3n - 1)(3n + 2)} = \frac{n}{6n + 4}$".

Base Case (n=1):

For $n=1$, the left side of the statement is the first term:

LHS $= \frac{1}{(3 \cdot 1 - 1)(3 \cdot 1 + 2)} = \frac{1}{(3 - 1)(3 + 2)} = \frac{1}{2 \cdot 5} = \frac{1}{10}$

The right side of the statement for $n=1$ is:

RHS $= \frac{1}{6 \cdot 1 + 4} = \frac{1}{6 + 4} = \frac{1}{10}$

Comparing LHS and RHS:

$\frac{1}{10} = \frac{1}{10}$

This is true. So, $P(1)$ is true.

Induction Hypothesis:

Assume that $P(k)$ is true for some natural number $k$. That is, assume:

$\frac{1}{2 \cdot 5} + \frac{1}{5 \cdot 8} + \frac{1}{8 \cdot 11} + \dots + \frac{1}{(3k - 1)(3k + 2)} = \frac{k}{6k + 4}$

Inductive Step (n=k+1):

We need to prove that $P(k+1)$ is true, i.e., we need to show that:

$\frac{1}{2 \cdot 5} + \frac{1}{5 \cdot 8} + \dots + \frac{1}{(3(k+1) - 1)(3(k+1) + 2)} = \frac{k+1}{6(k+1) + 4}$

Simplify the $(k+1)$-th term:

$(3(k+1) - 1)(3(k+1) + 2) = (3k + 3 - 1)(3k + 3 + 2) \ $$ = (3k + 2)(3k + 5)$

The statement for $n=k+1$ is:

$\frac{1}{2 \cdot 5} + \frac{1}{5 \cdot 8} + \dots + \frac{1}{(3k - 1)(3k + 2)} + \frac{1}{(3k + 2)(3k + 5)} = \frac{k+1}{6k + 10}$.

Consider the left side of the statement for $n=k+1$:

LHS $= \left( \frac{1}{2 \cdot 5} + \frac{1}{5 \cdot 8} + \dots + \frac{1}{(3k - 1)(3k + 2)} \right) + \frac{1}{(3k + 2)(3k + 5)}$

Using the induction hypothesis, we can replace the sum in the parenthesis with $\frac{k}{6k + 4}$:

LHS $= \frac{k}{6k + 4} + \frac{1}{(3k + 2)(3k + 5)}$

Factor the denominator $6k + 4 = 2(3k + 2)$:

LHS $= \frac{k}{2(3k + 2)} + \frac{1}{(3k + 2)(3k + 5)}$

Find a common denominator, which is $2(3k + 2)(3k + 5)$:

LHS $= \frac{k \cdot (3k + 5)}{2(3k + 2)(3k + 5)} + \frac{1 \cdot 2}{2(3k + 2)(3k + 5)}$

LHS $= \frac{k(3k + 5) + 2}{2(3k + 2)(3k + 5)}$

Expand and simplify the numerator:

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

Factor the quadratic numerator $3k^2 + 5k + 2$: $3k^2 + 3k + 2k + 2 = 3k(k+1) + 2(k+1) = (3k+2)(k+1)$

Substitute the factored numerator back into the LHS expression:

LHS $= \frac{(3k + 2)(k + 1)}{2(3k + 2)(3k + 5)}$

Cancel out the common factor $(3k+2)$ (since $k \in \mathbb{N}$, $3k+2$ is never zero):

LHS $= \frac{\cancel{(3k + 2)}(k + 1)}{2\cancel{(3k + 2)}(3k + 5)}$

LHS $= \frac{k + 1}{2(3k + 5)}$

Expand the denominator:

LHS $= \frac{k + 1}{6k + 10}$

This matches the right side of the statement for $n=k+1$ (RHS $= \frac{k+1}{6k + 10}$).

Since LHS = RHS, the statement $P(k+1)$ is true.

Conclusion:

By the Principle of Mathematical Induction, the statement $P(n): \frac{1}{2 \cdot 5} + \frac{1}{5 \cdot 8} + \frac{1}{8 \cdot 11} + \dots + \frac{1}{(3n - 1)(3n + 2)} = \frac{n}{6n + 4}$ is true for all natural numbers $n$.

Given:

The statement $P(n): \frac{1}{1 \cdot 2 \cdot 3} + \frac{1}{2 \cdot 3 \cdot 4} + \frac{1}{3 \cdot 4 \cdot 5} + \dots + \frac{1}{n(n + 1)(n + 2)} = \frac{n(n + 3)}{4 (n + 1)(n + 2)}$.

$n$ is a natural number ($n \in \{1, 2, 3, \dots \}$).

To Prove:

$P(n)$ is true for all natural numbers $n$, using the principle of mathematical induction.

Proof: Using Mathematical Induction

Let $P(n)$ be the statement "$\sum\limits_{r=1}^{n} \frac{1}{r(r+1)(r+2)} = \frac{n(n + 3)}{4 (n + 1)(n + 2)}$".

Base Case (n=1):

For $n=1$, the left side of the statement is the first term ($r=1$):

LHS $= \frac{1}{1 \cdot (1+1) \cdot (1+2)} = \frac{1}{1 \cdot 2 \cdot 3} = \frac{1}{6}$

The right side of the statement for $n=1$ is:

RHS $= \frac{1(1 + 3)}{4 (1 + 1) (1 + 2)} = \frac{1 \cdot 4}{4 \cdot 2 \cdot 3} = \frac{4}{24} = \frac{1}{6}$

Comparing LHS and RHS:

$\frac{1}{6} = \frac{1}{6}$

This is true. So, $P(1)$ is true.

Induction Hypothesis:

Assume that $P(k)$ is true for some natural number $k$. That is, assume:

$\frac{1}{1 \cdot 2 \cdot 3} + \frac{1}{2 \cdot 3 \cdot 4} + \dots + \frac{1}{k(k + 1)(k + 2)} = \frac{k(k + 3)}{4 (k + 1) (k + 2)}$

This is our induction hypothesis.

Inductive Step (n=k+1):

We need to prove that $P(k+1)$ is true, i.e., we need to show that:

$\frac{1}{1 \cdot 2 \cdot 3} + \frac{1}{2 \cdot 3 \cdot 4} + \dots + \frac{1}{(k+1)((k+1) + 1)((k+1) + 2)} = \frac{(k+1)((k+1) + 3)}{4 ((k+1) + 1)((k+1) + 2)}$

This simplifies to showing:

$\frac{1}{1 \cdot 2 \cdot 3} + \frac{1}{2 \cdot 3 \cdot 4} + \dots + \frac{1}{(k+1)(k + 2)(k + 3)} = \frac{(k+1)(k + 4)}{4 (k + 2)(k + 3)}$.

Consider the left side of the statement for $n=k+1$:

LHS $= \left( \frac{1}{1 \cdot 2 \cdot 3} + \dots + \frac{1}{k(k + 1)(k + 2)} \right) + \frac{1}{(k+1)(k + 2)(k + 3)}$

Using the induction hypothesis, we can replace the sum in the parenthesis with $\frac{k(k + 3)}{4 (k + 1) (k + 2)}$:

LHS $= \frac{k(k + 3)}{4 (k + 1) (k + 2)} + \frac{1}{(k+1)(k + 2)(k + 3)}$

To combine the fractions, find a common denominator, which is $4(k + 1)(k + 2)(k + 3)$:

LHS $= \frac{k(k + 3) \cdot (k + 3)}{4 (k + 1) (k + 2) \cdot (k + 3)} + \frac{1 \cdot 4}{4 \cdot (k+1)(k + 2)(k + 3)}$

LHS $= \frac{k(k + 3)^2 + 4}{4(k + 1)(k + 2)(k + 3)}$

Expand the numerator:

Numerator $= k(k^2 + 6k + 9) + 4 = k^3 + 6k^2 + 9k + 4$

So, LHS $= \frac{k^3 + 6k^2 + 9k + 4}{4(k + 1)(k + 2)(k + 3)}$.

Now, consider the numerator of the right side of the statement for $n=k+1$:

Target Numerator $= (k+1)(k+4)$

The target denominator for $n=k+1$ is $4(k+2)(k+3)$.

Let's see if the numerator $k^3 + 6k^2 + 9k + 4$ can be factored in a way that helps cancellation to reach the target RHS $\frac{(k+1)(k + 4)}{4 (k + 2)(k + 3)}$.

Let's try factoring $k^3 + 6k^2 + 9k + 4$. If $k=-1$ is a root, then $(-1)^3 + 6(-1)^2 + 9(-1) + 4 = -1 + 6 - 9 + 4 = 0$. So $(k+1)$ is a factor.

Using polynomial division or synthetic division, we find that $k^3 + 6k^2 + 9k + 4 = (k+1)(k^2 + 5k + 4)$.

Factor the quadratic $k^2 + 5k + 4 = (k+1)(k+4)$.

So, the numerator $k^3 + 6k^2 + 9k + 4 = (k+1)(k+1)(k+4) = (k+1)^2(k+4)$.

Substitute the factored numerator back into the LHS expression:

LHS $= \frac{(k+1)^2(k + 4)}{4(k + 1)(k + 2)(k + 3)}$

Cancel out one factor of $(k+1)$ from the numerator and denominator (since $k \in \mathbb{N}$, $k+1 \ne 0$):

LHS $= \frac{\cancel{(k+1)}(k + 1)(k + 4)}{4\cancel{(k + 1)}(k + 2)(k + 3)}$

LHS $= \frac{(k + 1)(k + 4)}{4(k + 2)(k + 3)}$

This is the right side of the statement for $n=k+1$ (RHS $= \frac{(k+1)(k + 4)}{4 (k + 2)(k + 3)}$).

Since LHS = RHS, the statement $P(k+1)$ is true.

Conclusion:

By the Principle of Mathematical Induction, the statement $P(n): \frac{1}{1 \cdot 2 \cdot 3} + \frac{1}{2 \cdot 3 \cdot 4} + \frac{1}{3 \cdot 4 \cdot 5} + \dots + \frac{1}{n(n + 1)(n + 2)} = \frac{n(n + 3)}{4 (n + 1)(n + 2)}$ is true for all natural numbers $n$.

Given:

The statement for the sum of a geometric series $a + ar + ar^2 + \dots + ar^{n-1} = \frac{a(r^n - 1)}{r - 1}$.

$n$ is a natural number ($n \in \{1, 2, 3, \dots \}$).

$a$ and $r$ are real numbers, with $r \ne 1$.

To Prove:

$a + ar + ar^2 + \dots + ar^{n-1} = \frac{a(r^n - 1)}{r - 1}$ for all natural numbers $n$, using the principle of mathematical induction.

Proof: Using Mathematical Induction

Let $P(n)$ be the statement "$a + ar + ar^2 + \dots + ar^{n-1} = \frac{a(r^n - 1)}{r - 1}$".

Base Case (n=1):

For $n=1$, the left side of the statement is the first term:

LHS $= a \cdot r^{1-1} = a \cdot r^0 = a \cdot 1 = a$

The right side of the statement for $n=1$ is:

RHS $= \frac{a(r^1 - 1)}{r - 1} = \frac{a(r - 1)}{r - 1}$

Since we are given $r \ne 1$, $r-1 \ne 0$, so we can cancel $(r-1)$:

RHS $= a$

Comparing LHS and RHS:

$a = a$

This is true. So, $P(1)$ is true.

Induction Hypothesis:

Assume that $P(k)$ is true for some natural number $k$. That is, assume:

$a + ar + ar^2 + \dots + ar^{k-1} = \frac{a(r^k - 1)}{r - 1}$

This is our induction hypothesis.

Inductive Step (n=k+1):

We need to prove that $P(k+1)$ is true, i.e., we need to show that:

$a + ar + ar^2 + \dots + ar^{(k+1)-1} = \frac{a(r^{k+1} - 1)}{r - 1}$

This simplifies to showing:

$a + ar + ar^2 + \dots + ar^{k} = \frac{a(r^{k+1} - 1)}{r - 1}$.

Consider the left side of the statement for $n=k+1$:

LHS $= (a + ar + ar^2 + \dots + ar^{k-1}) + ar^k$

Using the induction hypothesis, we can replace the sum in the parenthesis with $\frac{a(r^k - 1)}{r - 1}$:

LHS $= \frac{a(r^k - 1)}{r - 1} + ar^k$

Now, we simplify the expression. Find a common denominator $(r-1)$:

LHS $= \frac{a(r^k - 1)}{r - 1} + \frac{ar^k (r - 1)}{r - 1}$

LHS $= \frac{a(r^k - 1) + ar^k (r - 1)}{r - 1}$

Expand the numerator:

Numerator $= a \cdot r^k - a \cdot 1 + ar^k \cdot r - ar^k \cdot 1$

Numerator $= ar^k - a + ar^{k+1} - ar^k$

Combine like terms ($ar^k$ and $-ar^k$ cancel out):

Numerator $= ar^{k+1} - a$

Factor out $a$ from the numerator:

Numerator $= a(r^{k+1} - 1)$

Substitute the simplified numerator back into the LHS expression:

LHS $= \frac{a(r^{k+1} - 1)}{r - 1}$

This is the right side of the statement for $n=k+1$ (RHS $= \frac{a(r^{k+1} - 1)}{r - 1}$).

Since LHS = RHS, the statement $P(k+1)$ is true.

Conclusion:

By the Principle of Mathematical Induction, the statement $P(n): a + ar + ar^2 + \dots + ar^{n-1} = \frac{a(r^n - 1)}{r - 1}$ is true for all natural numbers $n$, provided $r \ne 1$.

Given:

The statement $\left( 1+\frac{3}{1} \right) \left( 1+\frac{5}{4} \right) \left( 1+\frac{7}{9} \right) … \left( 1+\frac{(2n \;+\; 1)}{n^{2}} \right) = (n \;+\; 1)^2$.

$n$ is a natural number ($n \in \{1, 2, 3, \dots \}$).

To Prove:

$\left( 1+\frac{3}{1} \right) \left( 1+\frac{5}{4} \right) \left( 1+\frac{7}{9} \right) … \left( 1+\frac{(2n \;+\; 1)}{n^{2}} \right) = (n \;+\; 1)^2$ for all natural numbers $n$, using the principle of mathematical induction.

Proof: Using Mathematical Induction

Let $P(n)$ be the statement "$\left( 1+\frac{3}{1} \right) \left( 1+\frac{5}{4} \right) \left( 1+\frac{7}{9} \right) … \left( 1+\frac{(2n \;+\; 1)}{n^{2}} \right) = (n \;+\; 1)^2$".

We can rewrite the general term in the product: $1 + \frac{2i + 1}{i^2} = \frac{i^2}{i^2} + \frac{2i + 1}{i^2} = \frac{i^2 + 2i + 1}{i^2} = \frac{(i+1)^2}{i^2}$.

So $P(n)$ can be written as $\prod\limits_{i=1}^{n} \left( 1+\frac{2i \;+\; 1}{i^{2}} \right) = \prod\limits_{i=1}^{n} \frac{(i+1)^{2}}{i^{2}} = (n \;+\; 1)^2$.

Base Case (n=1):

For $n=1$, the left side of the statement is the first term of the product (for $i=1$):

LHS $= \left( 1+\frac{2(1) \;+\; 1}{1^{2}} \right) = \left( 1+\frac{3}{1} \right) = 1 + 3 = 4$

The right side of the statement for $n=1$ is:

RHS $= (1 \;+\; 1)^2 = 2^2 = 4$

Comparing LHS and RHS:

$4 = 4$

This is true. So, $P(1)$ is true.

Induction Hypothesis:

Assume that $P(k)$ is true for some natural number $k$. That is, assume:

$\left( 1+\frac{3}{1} \right) \left( 1+\frac{5}{4} \right) \left( 1+\frac{7}{9} \right) … \left( 1+\frac{(2k \;+\; 1)}{k^{2}} \right) = (k \;+\; 1)^2$

This is our induction hypothesis.

Inductive Step (n=k+1):

We need to prove that $P(k+1)$ is true, i.e., we need to show that:

$\left( 1+\frac{3}{1} \right) \left( 1+\frac{5}{4} \right) \left( 1+\frac{7}{9} \right) … \left( 1+\frac{(2(k+1) \;+\; 1)}{(k+1)^{2}} \right) = ((k+1) \;+\; 1)^2 \ $$ = (k \;+\; 2)^2$.

The $(k+1)$-th term in the product is $\left( 1+\frac{(2(k+1) \;+\; 1)}{(k+1)^{2}} \right) = \left( 1+\frac{2k+2+1}{(k+1)^{2}} \right) = \left( 1+\frac{2k+3}{(k+1)^{2}} \right)$.

Consider the left side of the statement for $n=k+1$:

LHS $= \left[ \left( 1+\frac{3}{1} \right) \left( 1+\frac{5}{4} \right) … \left( 1+\frac{(2k \;+\; 1)}{k^{2}} \right) \right] \left( 1+\frac{(2(k+1) \;+\; 1)}{(k+1)^{2}} \right)$

Using the induction hypothesis, the product in the square brackets is equal to $(k+1)^2$:

LHS $= (k \;+\; 1)^2 \left( 1+\frac{2k+3}{(k+1)^{2}} \right)$

Simplify the term in the parenthesis:

$1+\frac{2k+3}{(k+1)^{2}} = \frac{(k+1)^2}{(k+1)^2} + \frac{2k+3}{(k+1)^{2}} = \frac{k^2+2k+1+2k+3}{(k+1)^2} = \frac{k^2+4k+4}{(k+1)^2} = \frac{(k+2)^2}{(k+1)^2}$

Substitute this back into the LHS expression:

LHS $= (k \;+\; 1)^2 \cdot \frac{(k+2)^2}{(k+1)^2}$

Since $k \in \mathbb{N}$, $k \ge 1$, so $(k+1)^2 \ne 0$. We can cancel the term $(k+1)^2$:

LHS $= \cancel{(k \;+\; 1)^2} \cdot \frac{(k+2)^2}{\cancel{(k+1)^2}}$

LHS $= (k+2)^2$

This is the right side of the statement for $n=k+1$ (RHS $= (k+2)^2$).

Since LHS = RHS, the statement $P(k+1)$ is true.

Conclusion:

By the Principle of Mathematical Induction, the statement $P(n): \left( 1+\frac{3}{1} \right) \left( 1+\frac{5}{4} \right) \left( 1+\frac{7}{9} \right) … \left( 1+\frac{(2n \;+\; 1)}{n^{2}} \right) = (n \;+\; 1)^2$ is true for all natural numbers $n$.

Given:

The statement $\left( 1+\frac{1}{1} \right) \left( 1+\frac{1}{2} \right) \left( 1+\frac{1}{3} \right) … \left( 1+\frac{1}{n} \right) = (n + 1)$.

$n$ is a natural number ($n \in \{1, 2, 3, \dots \}$).

To Prove:

$\left( 1+\frac{1}{1} \right) \left( 1+\frac{1}{2} \right) \left( 1+\frac{1}{3} \right) … \left( 1+\frac{1}{n} \right) = (n + 1)$ for all natural numbers $n$, using the principle of mathematical induction.

Proof: Using Mathematical Induction

Let $P(n)$ be the statement "$\left( 1+\frac{1}{1} \right) \left( 1+\frac{1}{2} \right) \left( 1+\frac{1}{3} \right) … \left( 1+\frac{1}{n} \right) = (n + 1)$".

We can rewrite the general term inside the parentheses: $1 + \frac{1}{i} = \frac{i}{i} + \frac{1}{i} = \frac{i+1}{i}$.

So $P(n)$ can be written as $\prod\limits_{i=1}^{n} \left( \frac{i+1}{i} \right) = n + 1$.

Base Case (n=1):

For $n=1$, the left side of the statement is the first term of the product (for $i=1$):

LHS $= \left( 1+\frac{1}{1} \right) = 1 + 1 = 2$

The right side of the statement for $n=1$ is:

RHS $= (1 + 1) = 2$

Comparing LHS and RHS:

$2 = 2$

This is true. So, $P(1)$ is true.

Induction Hypothesis:

Assume that $P(k)$ is true for some natural number $k$. That is, assume:

$\left( 1+\frac{1}{1} \right) \left( 1+\frac{1}{2} \right) \left( 1+\frac{1}{3} \right) … \left( 1+\frac{1}{k} \right) = (k + 1)$

This can be written as $\prod\limits_{i=1}^{k} \left( \frac{i+1}{i} \right) = k + 1$.

Inductive Step (n=k+1):

We need to prove that $P(k+1)$ is true, i.e., we need to show that:

$\left( 1+\frac{1}{1} \right) \left( 1+\frac{1}{2} \right) … \left( 1+\frac{1}{k} \right) \left( 1+\frac{1}{k+1} \right) = ((k+1) + 1) = k + 2$.

Consider the left side of the statement for $n=k+1$:

LHS $= \left[ \left( 1+\frac{1}{1} \right) \left( 1+\frac{1}{2} \right) … \left( 1+\frac{1}{k} \right) \right] \left( 1+\frac{1}{k+1} \right)$

Using the induction hypothesis, the product in the square brackets is equal to $(k+1)$:

LHS $= (k + 1) \left( 1+\frac{1}{k+1} \right)$

Simplify the term in the parenthesis:

$1+\frac{1}{k+1} = \frac{k+1}{k+1} + \frac{1}{k+1} = \frac{k+1+1}{k+1} = \frac{k+2}{k+1}$.

Substitute this back into the LHS expression:

LHS $= (k + 1) \cdot \left( \frac{k+2}{k+1} \right)$

Since $k$ is a natural number, $k \ge 1$, so $k+1 \ne 0$. We can cancel the term $(k+1)$:

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

LHS $= k + 2$

This is the right side of the statement for $n=k+1$ (RHS $= k + 2$).

Since LHS = RHS, the statement $P(k+1)$ is true.

Conclusion:

By the Principle of Mathematical Induction, the statement $P(n): \left( 1+\frac{1}{1} \right) \left( 1+\frac{1}{2} \right) \left( 1+\frac{1}{3} \right) … \left( 1+\frac{1}{n} \right) = (n + 1)$ is true for all natural numbers $n$.

Given:

The statement $1^2 + 3^2 + 5^2 + \dots + (2n – 1)^2 = \frac{n(2n – 1)(2n + 1)}{3}$.

$n$ is a natural number ($n \in \{1, 2, 3, \dots \}$).

To Prove:

$1^2 + 3^2 + 5^2 + \dots + (2n – 1)^2 = \frac{n(2n – 1)(2n + 1)}{3}$ for all natural numbers $n$, using the principle of mathematical induction.

Proof: Using Mathematical Induction

Let $P(n)$ be the statement "$1^2 + 3^2 + 5^2 + \dots + (2n – 1)^2 = \frac{n(2n – 1)(2n + 1)}{3}$".

Base Case (n=1):

For $n=1$, the left side of the statement is the first term:

LHS $= (2 \cdot 1 – 1)^2 = (2 – 1)^2 = 1^2 = 1$

The right side of the statement for $n=1$ is:

RHS $= \frac{1(2 \cdot 1 – 1)(2 \cdot 1 + 1)}{3} = \frac{1(2 – 1)(2 + 1)}{3} = \frac{1 \cdot 1 \cdot 3}{3} = \frac{3}{3} = 1$

Comparing LHS and RHS:

$1 = 1$

This is true. So, $P(1)$ is true.

Induction Hypothesis:

Assume that $P(k)$ is true for some natural number $k$. That is, assume:

$1^2 + 3^2 + 5^2 + \dots + (2k – 1)^2 = \frac{k(2k – 1)(2k + 1)}{3}$

Inductive Step (n=k+1):

We need to prove that $P(k+1)$ is true, i.e., we need to show that:

$1^2 + 3^2 + 5^2 + \dots + (2(k+1) – 1)^2 = \frac{(k+1)(2(k+1) – 1)(2(k+1) + 1)}{3}$

The $(k+1)$-th term is $(2(k+1) – 1)^2 = (2k + 2 – 1)^2 = (2k+1)^2$.

The right side for $n=k+1$ is $\frac{(k+1)(2k+2-1)(2k+2+1)}{3} = \frac{(k+1)(2k+1)(2k+3)}{3}$.

Consider the left side of the statement for $n=k+1$:

LHS $= (1^2 + 3^2 + 5^2 + \dots + (2k – 1)^2) + (2k+1)^2$

Using the induction hypothesis, we can replace the sum in the parenthesis:

LHS $= \frac{k(2k – 1)(2k + 1)}{3} + (2k+1)^2$

Find a common denominator:

LHS $= \frac{k(2k – 1)(2k + 1)}{3} + \frac{3(2k+1)^2}{3}$

Factor out the common term $(2k+1)$ from the numerator:

Numerator $= (2k+1) [k(2k-1) + 3(2k+1)]$

Expand the terms inside the square brackets:

$k(2k-1) + 3(2k+1) = (2k^2 - k) + (6k + 3) = 2k^2 + 5k + 3$

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 expression back into the numerator:

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

Substitute the numerator back into the LHS expression:

LHS $= \frac{(2k+1)(k+1)(2k+3)}{3}$

Rearrange the terms in the numerator:

LHS $= \frac{(k+1)(2k+1)(2k+3)}{3}$

This is the right side of the statement for $n=k+1$ (RHS $= \frac{(k+1)(2k+1)(2k+3)}{3}$).

Since LHS = RHS, the statement $P(k+1)$ is true.

Conclusion:

By the Principle of Mathematical Induction, the statement $P(n): 1^2 + 3^2 + 5^2 + \dots + (2n – 1)^2 = \frac{n(2n – 1)(2n + 1)}{3}$ is true for all natural numbers $n$.

Given:

The statement $\frac{1}{1 \cdot 4} + \frac{1}{4 \cdot 7} + \frac{1}{7 \cdot 10} + \dots + \frac{1}{(3n - 2)(3n + 1)} = \frac{n}{3n + 1}$.

$n$ is a natural number ($n \in \{1, 2, 3, \dots \}$).

To Prove:

$\frac{1}{1 \cdot 4} + \frac{1}{4 \cdot 7} + \frac{1}{7 \cdot 10} + \dots + \frac{1}{(3n - 2)(3n + 1)} = \frac{n}{3n + 1}$ for all natural numbers $n$, using the principle of mathematical induction.

Proof: Using Mathematical Induction

Let $P(n)$ be the statement "$\frac{1}{1 \cdot 4} + \frac{1}{4 \cdot 7} + \frac{1}{7 \cdot 10} + \dots + \frac{1}{(3n - 2)(3n + 1)} = \frac{n}{3n + 1}$".

Base Case (n=1):

For $n=1$, the left side of the statement is the first term:

LHS $= \frac{1}{(3 \cdot 1 - 2)(3 \cdot 1 + 1)} = \frac{1}{(3 - 2)(3 + 1)} = \frac{1}{1 \cdot 4} = \frac{1}{4}$

The right side of the statement for $n=1$ is:

RHS $= \frac{1}{(3 \cdot 1 + 1)} = \frac{1}{3 + 1} = \frac{1}{4}$

Comparing LHS and RHS:

$\frac{1}{4} = \frac{1}{4}$

This is true. So, $P(1)$ is true.

Induction Hypothesis:

Assume that $P(k)$ is true for some natural number $k$. That is, assume:

$\frac{1}{1 \cdot 4} + \frac{1}{4 \cdot 7} + \frac{1}{7 \cdot 10} + \dots + \frac{1}{(3k - 2)(3k + 1)} = \frac{k}{3k + 1}$

Inductive Step (n=k+1):

We need to prove that $P(k+1)$ is true, i.e., we need to show that:

$\frac{1}{1 \cdot 4} + \frac{1}{4 \cdot 7} + \dots + \frac{1}{(3(k+1) - 2)(3(k+1) + 1)} = \frac{k+1}{3(k+1) + 1}$

The $(k+1)$-th term in the sum is $\frac{1}{(3(k+1) - 2)(3(k+1) + 1)} = \frac{1}{(3k + 3 - 2)(3k + 3 + 1)} = \frac{1}{(3k + 1)(3k + 4)}$.

The statement for $n=k+1$ is:

$\frac{1}{1 \cdot 4} + \frac{1}{4 \cdot 7} + \dots + \frac{1}{(3k - 2)(3k + 1)} + \frac{1}{(3k + 1)(3k + 4)} = \frac{k+1}{3k + 4}$.

Consider the left side of the statement for $n=k+1$:

LHS $= \left( \frac{1}{1 \cdot 4} + \frac{1}{4 \cdot 7} + \dots + \frac{1}{(3k - 2)(3k + 1)} \right) + \frac{1}{(3k + 1)(3k + 4)}$

Using the induction hypothesis, we can replace the sum in the parenthesis with $\frac{k}{3k + 1}$:

LHS $= \frac{k}{3k + 1} + \frac{1}{(3k + 1)(3k + 4)}$

Find a common denominator, which is $(3k+1)(3k+4)$:

LHS $= \frac{k(3k + 4)}{(3k + 1)(3k + 4)} + \frac{1}{(3k + 1)(3k + 4)}$

Combine the terms in the numerator:

LHS $= \frac{k(3k + 4) + 1}{(3k + 1)(3k + 4)}$

Expand the numerator:

Numerator $= 3k^2 + 4k + 1$

Factor the quadratic numerator $3k^2 + 4k + 1$: $3k^2 + 3k + k + 1 = 3k(k+1) + 1(k+1) = (3k+1)(k+1)$.

Substitute the factored numerator back into the LHS expression:

LHS $= \frac{(3k + 1)(k + 1)}{(3k + 1)(3k + 4)}$

Cancel out the common factor $(3k+1)$ (since $k \in \mathbb{N}$, $3k+1$ is never zero):

LHS $= \frac{\cancel{(3k + 1)}(k + 1)}{\cancel{(3k + 1)}(3k + 4)}$

LHS $= \frac{k + 1}{3k + 4}$

This is the right side of the statement for $n=k+1$ (RHS $= \frac{k+1}{3k + 4}$).

Since LHS = RHS, the statement $P(k+1)$ is true.

Conclusion:

By the Principle of Mathematical Induction, the statement $P(n): \frac{1}{1 \cdot 4} + \frac{1}{4 \cdot 7} + \frac{1}{7 \cdot 10} + \dots + \frac{1}{(3n - 2)(3n + 1)} = \frac{n}{3n + 1}$ is true for all natural numbers $n$.

Given:

The statement $\frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \frac{1}{7 \cdot 9} + \dots + \frac{1}{(2n + 1)(2n + 3)} = \frac{n}{3(2n + 3)}$.

$n$ is a natural number ($n \in \{1, 2, 3, \dots \}$).

To Prove:

$\frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \frac{1}{7 \cdot 9} + \dots + \frac{1}{(2n + 1)(2n + 3)} = \frac{n}{3(2n + 3)}$ for all natural numbers $n$, using the principle of mathematical induction.

Proof: Using Mathematical Induction

Let $P(n)$ be the statement "$\frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \frac{1}{7 \cdot 9} + \dots + \frac{1}{(2n + 1)(2n + 3)} = \frac{n}{3(2n + 3)}$".

Base Case (n=1):

For $n=1$, the left side of the statement is the first term ($n=1$):

LHS $= \frac{1}{(2 \cdot 1 + 1)(2 \cdot 1 + 3)} = \frac{1}{(2 + 1)(2 + 3)} = \frac{1}{3 \cdot 5} = \frac{1}{15}$

The right side of the statement for $n=1$ is:

RHS $= \frac{1}{3(2 \cdot 1 + 3)} = \frac{1}{3(2 + 3)} = \frac{1}{3(5)} = \frac{1}{15}$

Comparing LHS and RHS:

$\frac{1}{15} = \frac{1}{15}$

This is true. So, $P(1)$ is true.

Induction Hypothesis:

Assume that $P(k)$ is true for some natural number $k$. That is, assume:

$\frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \frac{1}{7 \cdot 9} + \dots + \frac{1}{(2k + 1)(2k + 3)} = \frac{k}{3(2k + 3)}$

Inductive Step (n=k+1):

We need to prove that $P(k+1)$ is true, i.e., we need to show that:

$\frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \dots + \frac{1}{(2(k+1) + 1)(2(k+1) + 3)} = \frac{k+1}{3(2(k+1) + 3)}$

The $(k+1)$-th term in the sum is $\frac{1}{(2k + 2 + 1)(2k + 2 + 3)} = \frac{1}{(2k + 3)(2k + 5)}$.

The right side for $n=k+1$ is $\frac{k+1}{3(2k + 2 + 3)} = \frac{k+1}{3(2k + 5)}$.

Consider the left side of the statement for $n=k+1$:

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 induction hypothesis, we can replace the sum in the parenthesis:

LHS $= \frac{k}{3(2k + 3)} + \frac{1}{(2k + 3)(2k + 5)}$

Find a common denominator, which is $3(2k+3)(2k+5)$:

LHS $= \frac{k \cdot (2k + 5)}{3(2k + 3)(2k + 5)} + \frac{1 \cdot 3}{3(2k + 3)(2k + 5)}$

Combine the terms in the numerator:

LHS $= \frac{k(2k + 5) + 3}{3(2k + 3)(2k + 5)}$

Expand the numerator:

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

Factor the quadratic numerator $2k^2 + 5k + 3$. 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 numerator back into the LHS expression:

LHS $= \frac{(2k + 3)(k + 1)}{3(2k + 3)(2k + 5)}$

Cancel out the common factor $(2k+3)$ (since $k \in \mathbb{N}$, $2k+3$ is never zero):

LHS $= \frac{\cancel{(2k + 3)}(k + 1)}{3\cancel{(2k + 3)}(2k + 5)}$

LHS $= \frac{k + 1}{3(2k + 5)}$

This matches the right side of the statement for $n=k+1$ (RHS $= \frac{k+1}{3(2k + 5)}$).

Since LHS = RHS, the statement $P(k+1)$ is true.

Conclusion:

By the Principle of Mathematical Induction, the statement $P(n): \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \frac{1}{7 \cdot 9} + \dots + \frac{1}{(2n + 1)(2n + 3)} = \frac{n}{3(2n + 3)}$ is true for all natural numbers $n$.

Given:

The inequality $1 + 2 + 3 + \dots + n < \frac{1}{8} (2n + 1)^2$.

$n$ is a natural number ($n \in \{1, 2, 3, \dots \}$).

To Prove:

$1 + 2 + 3 + \dots + n < \frac{1}{8} (2n + 1)^2$ for all natural numbers $n$, using the principle of mathematical induction.

Proof: Using Mathematical Induction

Let $P(n)$ be the statement "$1 + 2 + 3 + \dots + n < \frac{1}{8} (2n + 1)^2$".

We know that the sum of the first $n$ natural numbers is given by the formula $1 + 2 + \dots + n = \frac{n(n+1)}{2}$. So, the statement $P(n)$ can be written as $\frac{n(n+1)}{2} < \frac{(2n+1)^2}{8}$.

Base Case (n=1):

For $n=1$, the left side of the inequality is:

LHS $= 1$

The right side of the inequality for $n=1$ is:

RHS $= \frac{1}{8} (2 \cdot 1 + 1)^2 = \frac{1}{8} (2 + 1)^2 = \frac{1}{8} (3)^2 = \frac{9}{8}$

Comparing LHS and RHS:

$1 < \frac{9}{8}$

This is true since $1 = \frac{8}{8}$ and $\frac{8}{8} < \frac{9}{8}$. So, $P(1)$ is true.

Induction Hypothesis:

Assume that $P(k)$ is true for some natural number $k$. That is, assume:

$1 + 2 + 3 + \dots + k < \frac{1}{8} (2k + 1)^2$

This can also be written as $\frac{k(k+1)}{2} < \frac{(2k+1)^2}{8}$.

Inductive Step (n=k+1):

We need to prove that $P(k+1)$ is true, i.e., we need to show that:

$1 + 2 + 3 + \dots + (k+1) < \frac{1}{8} (2(k+1) + 1)^2$

The right side of the inequality for $n=k+1$ simplifies to $\frac{1}{8} (2k + 2 + 1)^2 = \frac{1}{8} (2k + 3)^2$.

Consider the left side of the statement for $n=k+1$:

LHS $= (1 + 2 + 3 + \dots + k) + (k+1)$

Using the induction hypothesis, we know that $1 + 2 + 3 + \dots + k < \frac{1}{8} (2k + 1)^2$. Therefore, we can write:

LHS $< \frac{1}{8} (2k + 1)^2 + (k+1)$

Now, we want to show that $\frac{1}{8} (2k + 1)^2 + (k+1) \le \frac{1}{8} (2k + 3)^2$. If we can show this, then the strict inequality will follow from the induction hypothesis.

Let's simplify the expression $\frac{1}{8} (2k + 1)^2 + (k+1)$:

$\frac{1}{8} (2k + 1)^2 + (k+1) = \frac{(2k)^2 + 2(2k)(1) + 1^2}{8} + (k+1)$

$= \frac{4k^2 + 4k + 1}{8} + k + 1$

To combine the terms, find a common denominator (8):

$= \frac{4k^2 + 4k + 1}{8} + \frac{8(k+1)}{8}$

$= \frac{4k^2 + 4k + 1 + 8k + 8}{8}$

$= \frac{4k^2 + 12k + 9}{8}$

Now, let's consider the target RHS for $n=k+1$, which is $\frac{1}{8} (2k + 3)^2$. Expand $(2k+3)^2$:

$(2k + 3)^2 = (2k)^2 + 2(2k)(3) + 3^2 = 4k^2 + 12k + 9$

So, $\frac{1}{8} (2k + 3)^2 = \frac{4k^2 + 12k + 9}{8}$.

We found that $\frac{1}{8} (2k + 1)^2 + (k+1)$ is exactly equal to $\frac{4k^2 + 12k + 9}{8}$.

Therefore, we have:

$1 + 2 + 3 + \dots + (k+1) < \frac{1}{8} (2k + 1)^2 + (k+1) = \frac{4k^2 + 12k + 9}{8} \ $$ = \frac{(2k + 3)^2}{8}$

So, $1 + 2 + 3 + \dots + (k+1) < \frac{(2(k+1) + 1)^2}{8}$.

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

Conclusion:

By the Principle of Mathematical Induction, the statement $P(n): 1 + 2 + 3 + \dots + n < \frac{1}{8} (2n + 1)^2$ is true for all natural numbers $n$.

Given:

$n$ is a natural number ($n \in \{1, 2, 3, \dots \}$).

To Prove:

$n(n+1)(n+5)$ is a multiple of 3 for all natural numbers $n$.

Proof: Using Mathematical Induction

Let $P(n)$ be the statement "$n(n+1)(n+5)$ is divisible by 3".

Base Case (n=1):

For $n=1$, the expression is $1(1+1)(1+5)$.

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

Since $12 = 3 \times 4$, 12 is divisible by 3.

Thus, $P(1)$ is true.

Induction Hypothesis:

Assume that $P(k)$ is true for some natural number $k$.

That is, assume $k(k+1)(k+5)$ is divisible by 3.

This means there exists an integer $m$ such that $k(k+1)(k+5) = 3m$.

Inductive Step (n=k+1):

We need to prove that $P(k+1)$ is true, i.e., $(k+1)((k+1)+1)((k+1)+5)$ is divisible by 3.

Consider the expression for $n=k+1$:

$(k+1)(k+2)(k+6)$

We can rewrite the term $(k+6)$ as $(k+5) + 1$:

$(k+1)(k+2)(k+6) = (k+1)(k+2)((k+5) + 1)$

Distribute $(k+1)(k+2)$:

$(k+1)(k+2)(k+6) = (k+1)(k+2)(k+5) + (k+1)(k+2) \cdot 1$

Rearrange the terms to group with the induction hypothesis structure $k(k+1)(k+5)$:

$(k+1)(k+2)(k+5) = (k+1)(k+5)(k+2)$

$(k+1)(k+5)(k+2) = (k+1)(k+5)(k + 1 + 1)$

$= (k+1)(k+5)k + (k+1)(k+5)(1) + (k+1)(k+5)(1)$

$= k(k+1)(k+5) + (k+1)(k+5) + (k+1)(k+5)$

This expansion is complex. Let's use the expansion method again.

Consider the expression for $n=k+1$:

$(k+1)(k+2)(k+6)$

Expand this expression:

$(k+1)(k+2)(k+6) = (k^2 + 3k + 2)(k+6)$

$= k(k^2 + 3k + 2) + 6(k^2 + 3k + 2)$

$= k^3 + 3k^2 + 2k + 6k^2 + 18k + 12$

$= k^3 + 9k^2 + 20k + 12$

From the induction hypothesis, we know that $k(k+1)(k+5) = k(k^2 + 6k + 5) = k^3 + 6k^2 + 5k$ is divisible by 3.

Let's rewrite the expression for $n=k+1$ by grouping terms to isolate the expression from the induction hypothesis:

$(k^3 + 9k^2 + 20k + 12) = (k^3 + 6k^2 + 5k) + (3k^2 + 15k + 12)$

We can factor out 3 from the second group of terms:

$3k^2 + 15k + 12 = 3(k^2 + 5k + 4)$

So, $(k+1)(k+2)(k+6) = k(k+1)(k+5) + 3(k^2 + 5k + 4)$.

By the induction hypothesis, $k(k+1)(k+5)$ is divisible by 3.

The term $3(k^2 + 5k + 4)$ is clearly divisible by 3, because $k$ is an integer, which means $k^2$, $5k$, and $k^2 + 5k + 4$ are all integers. Therefore, $3 \times (\text{an integer})$ is divisible by 3.

Since both terms, $k(k+1)(k+5)$ and $3(k^2 + 5k + 4)$, are divisible by 3, their sum, $(k+1)(k+2)(k+6)$, is also divisible by 3.

Thus, $P(k+1)$ is true.

Conclusion:

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 $n$.

Given:

The statement $10^{2n - 1} + 1$ is divisible by 11.

$n$ is a natural number ($n \in \{1, 2, 3, \dots \}$).

To Prove:

$10^{2n - 1} + 1$ is divisible by 11 for all natural numbers $n$, using the principle of mathematical induction.

Proof: Using Mathematical Induction

Let $P(n)$ be the statement "$10^{2n - 1} + 1$ is divisible by 11".

Base Case (n=1):

For $n=1$, the expression is:

$10^{2(1) - 1} + 1 = 10^{2 - 1} + 1 = 10^1 + 1 = 10 + 1 = 11$

Since 11 is divisible by 11, $P(1)$ is true.

Induction Hypothesis:

Assume that $P(k)$ is true for some natural number $k$. That is, assume $10^{2k - 1} + 1$ is divisible by 11.

This means there exists an integer $m$ such that:

$10^{2k - 1} + 1 = 11m$

From this, we can write $10^{2k - 1} = 11m - 1$.

Inductive Step (n=k+1):

We need to prove that $P(k+1)$ is true, i.e., $10^{2(k+1) - 1} + 1$ is divisible by 11.

Consider the expression for $n=k+1$:

$10^{2(k+1) - 1} + 1 = 10^{2k + 2 - 1} + 1 = 10^{2k + 1} + 1$

Rewrite the term $10^{2k + 1}$ using properties of exponents:

$10^{2k + 1} = 10^{(2k - 1) + 2} = 10^{2k - 1} \cdot 10^2 = 100 \cdot 10^{2k - 1}$

Substitute this back into the expression for $n=k+1$:

$10^{2k + 1} + 1 = 100 \cdot 10^{2k - 1} + 1$

Now, use the induction hypothesis: $10^{2k - 1} = 11m - 1$. Substitute this into the equation:

$100 \cdot (11m - 1) + 1$

Expand the expression:

$1100m - 100 + 1$

Simplify the expression:

$1100m - 99$

Factor out 11 from the terms:

$11(100m - 9)$

Since $m$ is an integer (from the induction hypothesis), $100m - 9$ is also an integer.

Let $M = 100m - 9$. Then the expression is $11M$, which is divisible by 11.

Thus, $10^{2(k+1) - 1} + 1$ is divisible by 11. This shows that $P(k+1)$ is true.

Conclusion:

By the Principle of Mathematical Induction, the statement $P(n): 10^{2n - 1} + 1$ is divisible by 11 is true for all natural numbers $n$.

Given:

The statement $x^{2n} – y^{2n}$ is divisible by $x + y$.

$n$ is a natural number ($n \in \{1, 2, 3, \dots \}$).

We assume $x$ and $y$ are integers for the property of divisibility to be well-defined.

To Prove:

$x^{2n} – y^{2n}$ is divisible by $x + y$ for all natural numbers $n$, using the principle of mathematical induction.

Proof: Using Mathematical Induction

Let $P(n)$ be the statement "$x^{2n} - y^{2n}$ is divisible by $x + y$".

Base Case (n=1):

For $n=1$, the expression is $x^{2(1)} - y^{2(1)} = x^2 - y^2$.

Using the difference of squares formula, we have $x^2 - y^2 = (x - y)(x + y)$.

Since $(x^2 - y^2)$ can be written as a product of $(x+y)$ and $(x-y)$ (which is an integer because $x$ and $y$ are integers), $x^2 - y^2$ is divisible by $x+y$.

Thus, $P(1)$ is true.

Induction Hypothesis:

Assume that $P(k)$ is true for some arbitrary natural number $k$.

That is, assume $x^{2k} - y^{2k}$ is divisible by $x + y$.

This means there exists an integer $m$ such that $x^{2k} - y^{2k} = m(x+y)$.

Inductive Step (n=k+1):

We need to prove that $P(k+1)$ is true, i.e., $x^{2(k+1)} - y^{2(k+1)}$ is divisible by $x + y$.

Consider the expression for $n=k+1$:

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

Using properties of exponents, this can be written as:

$= x^{2k} \cdot x^2 - y^{2k} \cdot y^2$

To introduce the term from the induction hypothesis, $x^{2k} - y^{2k}$, we can add and subtract $x^{2k} y^2$:

$= x^{2k} x^2 - x^{2k} y^2 + x^{2k} y^2 - y^{2k} y^2$

Group the terms:

$= x^{2k}(x^2 - y^2) + y^2(x^{2k} - y^{2k})$

Now, use the fact that $x^2 - y^2 = (x-y)(x+y)$ and the induction hypothesis $x^{2k} - y^{2k} = m(x+y)$:

$= x^{2k} \cdot (x-y)(x+y) + y^2 \cdot m(x+y)$

Factor out the common term $(x+y)$:

$= (x+y) [x^{2k}(x-y) + y^2 m]$

Since $x, y, k$ are integers and $m$ is an integer (by the induction hypothesis), the term in the square brackets $[x^{2k}(x-y) + y^2 m]$ is an integer.

Therefore, $x^{2(k+1)} - y^{2(k+1)}$ is a product of $(x+y)$ and an integer, which means it is divisible by $(x+y)$.

Thus, $P(k+1)$ is true.

Conclusion:

By the Principle of Mathematical Induction, the statement "$x^{2n} - y^{2n}$ is divisible by $x + y$" is true for all natural numbers $n$, assuming $x$ and $y$ are integers.

Given:

The statement $3^{2n + 2} - 8n - 9$ is divisible by 8.

$n$ is a natural number ($n \in \{1, 2, 3, \dots \}$).

To Prove:

$3^{2n + 2} - 8n - 9$ is divisible by 8 for all natural numbers $n$, using the principle of mathematical induction.

Proof: Using Mathematical Induction

Let $P(n)$ be the statement "$3^{2n + 2} - 8n - 9$ is divisible by 8".

Base Case (n=1):

For $n=1$, evaluate the expression:

$3^{2(1) + 2} - 8(1) - 9 = 3^{2 + 2} - 8 - 9$

$= 3^4 - 17$

$= 81 - 17$

$= 64$

Since $64 = 8 \times 8$, 64 is divisible by 8.

Thus, $P(1)$ is true.

Induction Hypothesis:

Assume that $P(k)$ is true for some arbitrary natural number $k$. That is, assume:

$3^{2k + 2} - 8k - 9$ is divisible by 8.

This means there exists an integer $m$ such that:

$3^{2k + 2} - 8k - 9 = 8m$

From this, we can write $3^{2k + 2} = 8m + 8k + 9$.

Inductive Step (n=k+1):

We need to prove that $P(k+1)$ is true, i.e., $3^{2(k+1) + 2} - 8(k+1) - 9$ is divisible by 8.

Consider the expression for $n=k+1$:

$3^{2(k+1) + 2} - 8(k+1) - 9$

Simplify the exponent and the terms:

$= 3^{2k + 2 + 2} - (8k + 8) - 9$

$= 3^{2k + 4} - 8k - 8 - 9$

$= 3^{2k + 4} - 8k - 17$

Rewrite $3^{2k+4}$ using properties of exponents:

$3^{2k + 4} = 3^{2k + 2} \cdot 3^2 = 9 \cdot 3^{2k + 2}$

Substitute this back into the expression for $n=k+1$:

$= 9 \cdot 3^{2k + 2} - 8k - 17$

Now, use the induction hypothesis $3^{2k + 2} = 8m + 8k + 9$. Substitute this into the equation:

$= 9 (8m + 8k + 9) - 8k - 17$

Expand the expression:

$= 9 \cdot 8m + 9 \cdot 8k + 9 \cdot 9 - 8k - 17$

$= 72m + 72k + 81 - 8k - 17$

Group terms with $k$ and constant terms:

$= 72m + (72k - 8k) + (81 - 17)$

$= 72m + 64k + 64$

Factor out 8 from the terms:

$= 8(9m + 8k + 8)$

Since $m$ is an integer (from the induction hypothesis) and $k$ is a natural number, the term inside the parenthesis $(9m + 8k + 8)$ is an integer.

Let $M = 9m + 8k + 8$. Then the expression is $8M$, which is a multiple of 8.

Therefore, $3^{2(k+1) + 2} - 8(k+1) - 9$ is divisible by 8. This shows that $P(k+1)$ is true.

Conclusion:

By the Principle of Mathematical Induction, the statement $P(n): 3^{2n + 2} - 8n - 9$ is divisible by 8 is true for all natural numbers $n$.

Given:

The statement $41^n – 14^n$ is a multiple of 27.

$n$ is a natural number ($n \in \{1, 2, 3, \dots \}$).

To Prove:

$41^n – 14^n$ is divisible by 27 for all natural numbers $n$, using the principle of mathematical induction.

Proof: Using Mathematical Induction

Let $P(n)$ be the statement "$41^n - 14^n$ is divisible by 27".

Base Case (n=1):

For $n=1$, the expression is:

$41^1 - 14^1 = 41 - 14 = 27$

Since 27 is divisible by 27, $P(1)$ is true.

Induction Hypothesis:

Assume that $P(k)$ is true for some arbitrary natural number $k$. That is, assume $41^k - 14^k$ is divisible by 27.

This means there exists an integer $m$ such that:

$41^k - 14^k = 27m$

From this, we can write $41^k = 14^k + 27m$.

Inductive Step (n=k+1):

We need to prove that $P(k+1)$ is true, i.e., $41^{k+1} - 14^{k+1}$ is divisible by 27.

Consider the expression for $n=k+1$:

$41^{k+1} - 14^{k+1}$

Rewrite using properties of exponents:

$= 41 \cdot 41^k - 14 \cdot 14^k$

Use the induction hypothesis: $41^k = 14^k + 27m$. Substitute this into the equation:

$= 41 (14^k + 27m) - 14 \cdot 14^k$

Expand the expression:

$= 41 \cdot 14^k + 41 \cdot 27m - 14 \cdot 14^k$

Group the terms containing $14^k$:

$= (41 \cdot 14^k - 14 \cdot 14^k) + 41 \cdot 27m$

Factor out $14^k$ from the first group:

$= 14^k (41 - 14) + 41 \cdot 27m$

Simplify the term in the parenthesis:

$= 14^k (27) + 41 \cdot 27m$

Factor out 27 from both terms:

$= 27 (14^k + 41m)$

Since $k$ is a natural number, $14^k$ is an integer. By the induction hypothesis, $m$ is an integer. Therefore, the term inside the parenthesis $(14^k + 41m)$ is an integer.

Let $M = 14^k + 41m$. Then $41^{k+1} - 14^{k+1} = 27M$, which is a multiple of 27.

Thus, $41^{k+1} - 14^{k+1}$ is divisible by 27. This shows that $P(k+1)$ is true.

Conclusion:

By the Principle of Mathematical Induction, the statement $P(n): 41^n - 14^n$ is divisible by 27 is true for all natural numbers $n$.

Given:

The inequality $(2n + 7) < (n + 3)^2$.

$n$ is a natural number ($n \in \{1, 2, 3, \dots \}$).

To Prove:

$(2n + 7) < (n + 3)^2$ for all natural numbers $n$, using the principle of mathematical induction.

Proof: Using Mathematical Induction

Let $P(n)$ be the statement "$(2n + 7) < (n + 3)^2$".

Base Case (n=1):

For $n=1$, the left side of the inequality is:

LHS $= 2(1) + 7 = 2 + 7 = 9$

The right side of the inequality for $n=1$ is:

RHS $= (1 + 3)^2 = 4^2 = 16$

Comparing LHS and RHS:

$9 < 16$

This is true. So, $P(1)$ is true.

Induction Hypothesis:

Assume that $P(k)$ is true for some natural number $k$. That is, assume:

$(2k + 7) < (k + 3)^2$

$(2k + 7) < k^2 + 6k + 9$

Inductive Step (n=k+1):

We need to prove that $P(k+1)$ is true, i.e., we need to show that $(2(k+1) + 7) < ((k+1) + 3)^2$.

Simplify the inequality we need to show:

$(2k + 2 + 7) < (k + 4)^2$

$(2k + 9) < k^2 + 8k + 16$

Consider the left side of the statement for $n=k+1$:

LHS$_{k+1} = 2(k+1) + 7 = 2k + 2 + 7 = 2k + 9$.

From the induction hypothesis, we have $2k + 7 < (k+3)^2$.

Add 2 to both sides of this inequality:

$2k + 7 + 2 < (k+3)^2 + 2$

$2k + 9 < k^2 + 6k + 9 + 2$

$2k + 9 < k^2 + 6k + 11$

Now, consider the right side of the inequality for $n=k+1$:

RHS$_{k+1} = (k+4)^2 = k^2 + 8k + 16$.

We need to show that $2k + 9 < k^2 + 8k + 16$.

We know that $2k + 9 < k^2 + 6k + 11$. Let's compare $k^2 + 6k + 11$ with $k^2 + 8k + 16$.

Consider the difference: $(k^2 + 8k + 16) - (k^2 + 6k + 11) = 2k + 5$.

Since $k$ is a natural number, $k \ge 1$. Thus, $2k \ge 2$, and $2k + 5 \ge 2 + 5 = 7$.

Since $7 > 0$, the difference is positive: $(k^2 + 8k + 16) - (k^2 + 6k + 11) > 0$.

This means $k^2 + 6k + 11 < k^2 + 8k + 16$.

Combining the inequalities:

$2k + 9 < k^2 + 6k + 11$ (from adding 2 to the IH)

$k^2 + 6k + 11 < k^2 + 8k + 16$ (as shown above)

By the transitive property of inequalities, we have:

$2k + 9 < k^2 + 8k + 16$

$(2(k+1) + 7) < ((k+1) + 3)^2$

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

Conclusion:

By the Principle of Mathematical Induction, the statement $P(n): (2n + 7) < (n + 3)^2$ is true for all natural numbers $n$.