Binomial Theorem

Development and Statement of Binomial Theorem for Positive Integral Index

Expanding Binomials

A binomial is a simple algebraic expression that consists of exactly two terms, connected by an addition or subtraction sign, such as $(a+b)$, $(x-y)$, $(2p+3q)$, etc. When a binomial is raised to a positive integer power, like $(a+b)^2$ or $(x-y)^5$, we can expand it by multiplying the binomial by itself the required number of times. For small powers, this is straightforward:

• $(a+b)^1 = a+b$
• $(a+b)^2 = (a+b)(a+b) = a^2 + 2ab + b^2$
• $(a+b)^3 = (a+b)(a^2 + 2ab + b^2) = a^3 + 3a^2b + 3ab^2 + b^3$

However, as the power increases, performing repeated multiplication becomes very tedious and prone to errors. The Binomial Theorem provides a systematic and efficient formula for expanding any binomial $(a+b)$ raised to any positive integer power $n$.

Development of the Binomial Theorem

Let's look closely at the expanded forms of $(a+b)^n$ for the first few non-negative integer values of $n$ to observe the patterns:

• $n=0$: $(a+b)^0 = 1$ (assuming $a+b \neq 0$)
• $n=1$: $(a+b)^1 = a + b$
• $n=2$: $(a+b)^2 = a^2 + 2ab + b^2$
• $n=3$: $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$
• $n=4$: $(a+b)^4 = (a+b)(a+b)^3 = (a+b)(a^3 + 3a^2b + 3ab^2 + b^3)$ $$ = a(a^3 + 3a^2b + 3ab^2 + b^3) + b(a^3 + 3a^2b + 3ab^2 + b^3) $$ $$ = a^4 + 3a^3b + 3a^2b^2 + ab^3 + a^3b + 3a^2b^2 + 3ab^3 + b^4 $$ $$ = a^4 + (3+1)a^3b + (3+3)a^2b^2 + (1+3)ab^3 + b^4 $$ $$ = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4 $$

Let's list the expansions and identify the patterns in the terms and coefficients:

From these expansions, we can observe several important patterns:

1. The number of terms in the expansion of $(a+b)^n$ is always $n+1$.
2. In each term of the expansion, the sum of the exponents of $a$ and $b$ is always equal to $n$. (e.g., for $n=4$, terms have $a^4b^0$, $a^3b^1$, $a^2b^2$, $a^1b^3$, $a^0b^4$; in each case, $4+0=4, 3+1=4$, etc.).
3. As we move from left to right across the terms, the powers of $a$ decrease by 1 in each successive term (starting from $a^n$ down to $a^0$), and the powers of $b$ increase by 1 (starting from $b^0$ up to $b^n$).
4. The coefficients of the terms follow a distinct pattern. Comparing the list of coefficients (1; 1,1; 1,2,1; 1,3,3,1; 1,4,6,4,1) with Pascal's triangle, we see they are identical to the entries in the $n$-th row of Pascal's triangle (counting the top row as row 0).

The numbers in Pascal's triangle are precisely the binomial coefficients $\binom{n}{r} = C(n, r) = \frac{n!}{r!(n-r)!}$. Let's confirm this for a few rows:

• $n=0$: $\binom{0}{0} = \frac{0!}{0!0!} = \frac{1}{1 \times 1} = 1$.
• $n=1$: $\binom{1}{0} = \frac{1!}{0!1!} = \frac{1}{1 \times 1} = 1$, $\binom{1}{1} = \frac{1!}{1!0!} = \frac{1}{1 \times 1} = 1$. Coefficients are 1, 1.
• $n=2$: $\binom{2}{0} = \frac{2!}{0!2!} = 1$, $\binom{2}{1} = \frac{2!}{1!1!} = 2$, $\binom{2}{2} = \frac{2!}{2!0!} = 1$. Coefficients are 1, 2, 1.
• $n=3$: $\binom{3}{0} = \frac{3!}{0!3!} = 1$, $\binom{3}{1} = \frac{3!}{1!2!} = 3$, $\binom{3}{2} = \frac{3!}{2!1!} = 3$, $\binom{3}{3} = \frac{3!}{3!0!} = 1$. Coefficients are 1, 3, 3, 1.
• $n=4$: $\binom{4}{0} = \frac{4!}{0!4!} = 1$, $\binom{4}{1} = \frac{4!}{1!3!} = 4$, $\binom{4}{2} = \frac{4!}{2!2!} = 6$, $\binom{4}{3} = \frac{4!}{3!1!} = 4$, $\binom{4}{4} = \frac{4!}{4!0!} = 1$. Coefficients are 1, 4, 6, 4, 1.

The coefficient of the term with $a^{n-r}b^r$ (or $a^r b^{n-r}$) in the expansion of $(a+b)^n$ is $\binom{n}{r}$.

Statement of the Binomial Theorem for Positive Integral Index

Based on the patterns observed in the expansions and the connection to binomial coefficients, the Binomial Theorem for a positive integer exponent $n$ can be formally stated.

Statement of the Binomial Theorem:

For any real numbers $a$ and $b$, and any positive integer $n$, the expansion of $(a+b)^n$ is given by the sum of terms where the coefficient of $a^{n-r}b^r$ is $\binom{n}{r}$:

Using summation notation ($\sum$), which provides a concise way to represent sums of a sequence of terms, the Binomial Theorem can be written as:

In this formula:

• $n$ is the positive integer exponent.
• $r$ is an index variable that starts from $0$ and goes up to $n$. Each value of $r$ corresponds to a term in the expansion.
• $\binom{n}{r}$ (read as "n choose r") is the binomial coefficient, calculated using the formula $\frac{n!}{r!(n-r)!}$. These are the coefficients from the $(n+1)$-th row of Pascal's triangle (if we count the top '1' as row 1, or the $n$-th row if we count the top '1' as row 0).
• $a^{n-r} b^r$ represents the variable part of each term. The power of $a$ decreases from $n$ (when $r=0$) to $0$ (when $r=n$), and the power of $b$ increases from $0$ (when $r=0$) to $n$ (when $r=n$). Note that $(n-r) + r = n$, so the sum of the exponents is always $n$.

The terms in the expansion are generated by letting $r$ take integer values from $0$ to $n$. For $r=0$, the term is $\binom{n}{0} a^n b^0$. For $r=1$, it's $\binom{n}{1} a^{n-1} b^1$, and so on, until $r=n$, giving the term $\binom{n}{n} a^0 b^n$.

Proof (Outline using Induction)

The Binomial Theorem for a positive integral index $n$ can be formally proven using the Principle of Mathematical Induction on the exponent $n$.

1. Basis Step: Show that the theorem is true for the smallest positive integer, $n=1$. $$ (a+b)^1 = a+b $$

   According to the formula $\sum_{r=0}^{1} \binom{1}{r} a^{1-r} b^r$: For $r=0$: $\binom{1}{0} a^{1-0} b^0 = 1 \cdot a^1 \cdot 1 = a$. For $r=1$: $\binom{1}{1} a^{1-1} b^1 = 1 \cdot a^0 \cdot b^1 = 1 \cdot 1 \cdot b = b$. The sum is $a+b$. So, $(a+b)^1 = a+b$. The theorem is true for $n=1$.

2. Inductive Hypothesis: Assume that the theorem is true for some arbitrary positive integer $k$.

   Assume that $(a+b)^k = \sum_{r=0}^{k} \binom{k}{r} a^{k-r} b^r$ is true for some $k \ge 1$.

3. Inductive Step: We need to prove that the theorem is true for $n=k+1$, i.e., that $(a+b)^{k+1} = \sum_{r=0}^{k+1} \binom{k+1}{r} a^{k+1-r} b^r$.

   Consider $(a+b)^{k+1} = (a+b) \cdot (a+b)^k$.

   Using the inductive hypothesis for $(a+b)^k$:

   $$ (a+b)^{k+1} = (a+b) \left(\sum_{r=0}^{k} \binom{k}{r} a^{k-r} b^r\right) $$ $$ (a+b)^{k+1} = a \left(\sum_{r=0}^{k} \binom{k}{r} a^{k-r} b^r\right) + b \left(\sum_{r=0}^{k} \binom{k}{r} a^{k-r} b^r\right) $$ $$ (a+b)^{k+1} = \sum_{r=0}^{k} \binom{k}{r} a^{k-r+1} b^r + \sum_{r=0}^{k} \binom{k}{r} a^{k-r} b^{r+1} $$

   Manipulate these sums and combine terms with the same powers of $a$ and $b$ (i.e., $a^{k+1-m}b^m$ for some $m$). This involves shifting the index in the second sum and using the property of binomial coefficients $\binom{k}{r} + \binom{k}{r-1} = \binom{k+1}{r}$ (Pascal's Identity). After careful algebraic manipulation, the sum can be shown to equal $\sum_{r=0}^{k+1} \binom{k+1}{r} a^{k+1-r} b^r$.

4. Conclusion: Since the Basis Step and the Inductive Step hold, by the Principle of Mathematical Induction, the Binomial Theorem is true for all positive integers $n$.

Applying the Binomial Theorem

Let's use the Binomial Theorem formula to expand binomial expressions.

Answer:

Answer:

The Binomial Theorem provides a systematic method for expanding $(a+b)^n$ for any positive integer $n$. It is a cornerstone result in algebra and has wide applications in probability, statistics, and other areas of mathematics.

General Term of a Binomial Expansion

Finding a Specific Term

The Binomial Theorem provides the complete expanded form of $(a+b)^n$ as a sum of $n+1$ terms. However, in many problems, we don't need the entire expansion. We might only be interested in the value of a particular term, such as the 5th term, or a term with a specific power of $a$ or $b$, or even the term that does not contain the variable at all (the constant term).

The formula for the general term of a binomial expansion allows us to directly calculate any desired term based on its position or properties, without having to compute all the preceding terms.

Formula for the General Term

Let's revisit the Binomial Theorem expansion of $(a+b)^n$ for a positive integer $n$:

This expansion is a sum of terms, indexed by $r$, starting from $r=0$ up to $r=n$. Let's look at how the term number relates to the index $r$:

• When $r=0$, the term is $\binom{n}{0} a^n b^0$. This is the 1st term of the expansion.
• When $r=1$, the term is $\binom{n}{1} a^{n-1} b^1$. This is the 2nd term of the expansion.
• When $r=2$, the term is $\binom{n}{2} a^{n-2} b^2$. This is the 3rd term of the expansion.
• ...
• In general, when the index is $r$, the term is the $(r+1)$-th term of the expansion.

So, the formula for the $(r+1)$-th term, denoted by $T_{r+1}$, in the expansion of $(a+b)^n$ is given by the general form of the terms in the summation:

Here:

• $T_{r+1}$ is the $(r+1)$-th term in the expansion.
• $n$ is the exponent of the binomial.
• $a$ is the first term of the binomial.
• $b$ is the second term of the binomial (including its sign).
• $r$ is an integer index that determines the term number. $r$ can range from $0$ to $n$. If you are looking for the $k$-th term (where $k$ is the term number, $1 \le k \le n+1$), you set $r+1 = k$, which means $r = k-1$. Then substitute $r=k-1$ into the general term formula to find the $k$-th term.
• $\binom{n}{r}$ is the binomial coefficient $\frac{n!}{r!(n-r)!}$.

Examples Finding a Specific Term

Let's use the general term formula to find specific terms in binomial expansions.

Answer:

Answer:

The general term formula is a powerful tool that allows us to find any specific term in a binomial expansion without needing to compute the entire series, which is particularly useful for larger exponents or when the term's position or properties are specified.

Middle Term(s) of a Binomial Expansion

Locating the Central Term(s)

The expansion of a binomial $(a+b)^n$ has a total of $n+1$ terms. When we look at the sequence of terms in the expansion, we might be interested in finding the term(s) that are located in the exact middle of the sequence. The position of the middle term, or terms, depends on whether the total number of terms $(n+1)$ is odd or even.

Finding the Middle Term(s)

To find the middle term(s) in the expansion of $(a+b)^n$, we first determine the total number of terms, which is $n+1$. Then, we consider two cases based on the parity (whether it's even or odd) of the exponent $n$.

Case 1: When n is Even

If the exponent $n$ is an even positive integer (e.g., $n=2, 4, 6, 8, \ldots$), then the total number of terms in the expansion, $n+1$, is an odd number (e.g., $2+1=3, 4+1=5, 6+1=7, 8+1=9, \ldots$). When there is an odd number of terms in a sequence, there is exactly one term located in the middle.

The position of the single middle term is found by taking the total number of terms, adding 1, and dividing by 2. Since the total number of terms is $n+1$, the position of the middle term is $\frac{(n+1)+1}{2} = \frac{n+2}{2} = \frac{n}{2} + 1$.

Example: If $n=4$, the number of terms is $4+1=5$. The terms are $T_1, T_2, T_3, T_4, T_5$. The middle term is the 3rd term. The position formula gives $\frac{4}{2} + 1 = 2 + 1 = 3$.

To find the value of this middle term, we use the formula for the $(r+1)$-th term, $T_{r+1} = \binom{n}{r} a^{n-r} b^r$. Since the middle term is at position $\frac{n}{2} + 1$, we set $r+1 = \frac{n}{2} + 1$. Subtracting 1 from both sides gives $r = \frac{n}{2}$.

Substitute this value of $r$ into the general term formula:

Example

Answer:

Case 2: When n is Odd

If the exponent $n$ is an odd positive integer (e.g., $n=1, 3, 5, 7, \ldots$), then the total number of terms in the expansion, $n+1$, is an even number (e.g., $1+1=2, 3+1=4, 5+1=6, 7+1=8, \ldots$). When there is an even number of terms in a sequence, there are two terms located in the middle.

The positions of the two middle terms are found by dividing the total number of terms by 2 and taking that result and the next integer as the positions. Since the total number of terms is $n+1$, the positions of the two middle terms are $\frac{n+1}{2}$ and $\frac{n+1}{2} + 1$.

Example: If $n=3$, the number of terms is $3+1=4$. The terms are $T_1, T_2, T_3, T_4$. The middle terms are the 2nd and 3rd terms. The position formulas give $\frac{3+1}{2} = \frac{4}{2} = 2$ and $2+1 = 3$.

To find the value of these middle terms, we use the formula $T_{r+1} = \binom{n}{r} a^{n-r} b^r$.

• The first middle term is at position $\frac{n+1}{2}$. We set $r+1 = \frac{n+1}{2}$, so $r = \frac{n+1}{2} - 1 = \frac{n+1-2}{2} = \frac{n-1}{2}$. The term is $T_{\frac{n+1}{2}}$.
• The second middle term is at position $\frac{n+1}{2} + 1$. We set $r+1 = \frac{n+1}{2} + 1$, so $r = \frac{n+1}{2} + 1 - 1 = \frac{n+1}{2}$. The term is $T_{\frac{n+1}{2} + 1}$.

Substitute these values of $r$ into the general term formula:

Note that $\binom{n}{(n-1)/2} = \binom{n}{n - (n-1)/2} = \binom{n}{(2n-n+1)/2} = \binom{n}{(n+1)/2}$. The binomial coefficients for the two middle terms when $n$ is odd are always equal.

Example

Answer:

Identifying the middle term(s) is a common task that utilizes the general term formula of the binomial expansion. It's important to first determine whether $n$ is even or odd to know how many middle terms exist and what their positions are.