Binomial Theorem

In this chapter we study the expansion of powers of a binomial expression such as $(a+b)^n$, where $n$ is a non negative integer.
We will discover patterns in the coefficients using Pascal's triangle, and then state and use the Binomial Theorem.

Binomial Expansion for $n\equal 2$ and $n\equal 3$

Proposition Perfect Squares Expansion

The square of a sum and the square of a difference can be written as:$$(a+b)^2=a^2+2ab+b^2 \quad\text{and}\quad (a-b)^2=a^2-2ab+b^2.$$

Proof

$$\begin{aligned}(a+b)^2 &= (a+b)(a+b) &&\text{(definition of a square)} \\ &= a(a+b)+b(a+b) &&\text{(distributive law)} \\ &= a^2+ab+ab+b^2 &&\text{(expanding)} \\ &= a^2+2ab+b^2 &&\text{(combining like terms)}.\end{aligned}$$Similarly,$$\begin{aligned}(a-b)^2 &= (a-b)(a-b) &&\text{(definition of a square)} \\ &= a(a-b)-b(a-b) &&\text{(distributive law)} \\ &= a^2-ab-ab+b^2 &&\text{(expanding)} \\ &= a^2-2ab+b^2 &&\text{(combining like terms)}.\end{aligned}$$

Example

Expand and simplify $(x+2)^2$.

Answer

Using the formula $(a+b)^2=a^2+2ab+b^2$ with $a=x$ and $b=2$:$$\begin{aligned}(\textcolor{colordef}{x}+\textcolor{colorprop}{2})^{2} &=\textcolor{colordef}{x}^{2}+2 \times \textcolor{colordef}{x} \times \textcolor{colorprop}{2}+\textcolor{colorprop}{2}^{2} \\ &=x^{2}+4 x+4.\end{aligned}$$So $(x+2)^2=x^2+4x+4$.

Proposition Perfect Cube Expansion

Proof

$$\begin{aligned}(a+b)^3 &= (a+b)(a+b)(a+b) &&\text {(cube definition)} \\ &= (a^2+2ab+b^2)(a+b) &&\text {(using the square expansion)} \\ &= (a^2+2ab+b^2)a + (a^2+2ab+b^2)b &&\text {(expanding)} \\ &= a^3+2a^2b+ab^2 + a^2b+2ab^2+b^3 &&\text {(distributive law)} \\ &= a^3+3a^2b+3ab^2+b^3 &&\text {(combining)}\\ \end{aligned}$$

Example

Expand and simplify $(x+2)^3$

Answer

In the perfect cube expansion, we substitute $a=x$ and $b=2$:$$\begin{aligned}(\textcolor{colordef}{x}+\textcolor{colorprop}{2})^{3}&= \textcolor{colordef}{x}^{3} + 3 \times \textcolor{colordef}{x}^2 \times \textcolor{colorprop}{2} + 3 \times \textcolor{colordef}{x} \times \textcolor{colorprop}{2}^2 + \textcolor{colorprop}{2}^{3} \\ &= x^{3} + 6x^2 + 12x + 8\end{aligned}$$

Pascal's Triangle

Discover

Consider the powers of $(a+b)$:$$\begin{array}{rcccccccccc}(a+b)^1 &=& & & a &+& b\\ (a+b)^2 &=& & 1a^2 &+& 2ab &+& 1b^2\\ (a+b)^3 &=& 1a^3 &+& 3a^2b &+& 3ab^2 &+& 1b^3\\ (a+b)^4 &=& 1a^4 &+& 4a^3b &+& 6a^2b^2 &+& 4ab^3 &+& 1b^4\\ \end{array}$$Now, when we list only the coefficients of the terms, we get Pascal's triangle:$$\begin{array}{ccccccccccc}&&&&1&&1&&&&\text{row 1}\\ &&&1&&2&&1&&&\text{row 2}\\ &&1&&3&&3&&1&&\text{row 3}\\ &1&&4&&6&&4&&1&\text{row 4}\\ \end{array}$$We observe the following pattern.

Definition Pascal's Triangle

The values at the ends of each row are always $1$.
Each interior value is found by adding the two values diagonally above it.

Example

Find the 5th row of Pascal's triangle.

Answer

$$\begin{array}{cccccccccccc}&&&&1&&1&&&&&\text{row 1}\\ &&&1&&2&&1&&&&\text{row 2}\\ &&1&&3&&3&&1&&&\text{row 3}\\ &1&&4&&6&&4&&1&&\text{row 4}\\ 1&&5&&10&&10&&5&&1&\text{row 5}\end{array}$$So the 5th row is $1, 5, 10, 10, 5, 1$.

Proposition Binomial Expansion

For the binomial expansion of $(a+b)^n$ where $n \in \mathbb{N}$:

As we look from left to right across the expansion, the powers of $a$ decrease by $1$, while the powers of $b$ increase by $1$.
The sum of the powers of $a$ and $b$ in each term of the expansion is $n$.
The number of terms in the expansion is $n+1$.
The coefficients of the terms are row $n$ of Pascal's triangle.

Example

Find the binomial expansion of $(a+b)^5$.

Answer

From the 5th row of Pascal's triangle$$\begin{array}{cccccccccccc}&&&&1&&1&&&&&\text{row 1}\\ &&&1&&2&&1&&&&\text{row 2}\\ &&1&&3&&3&&1&&&\text{row 3}\\ &1&&4&&6&&4&&1&&\text{row 4}\\ 1&&5&&10&&10&&5&&1&\text{row 5}\end{array}$$we get$$(a+b)^5= a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5.$$

Binomial Theorem

Definition Factorial

For any positive integer $n$, $n!$ (read as "$n$ factorial") is the product of the first $n$ positive integers: $$n! = n \times (n-1) \times \dots \times 2 \times 1.$$ By convention, we define $0! = 1$.

Example

Calculate $4!$.

Answer

$\begin{aligned}[t]4! &= 4 \times 3 \times 2 \times 1\\ &= 24\end{aligned}$

Definition Binomial Coefficient

For any integers $n \geqslant p \geqslant 0$, the binomial coefficient $\binom{n}{p}$ is defined as $$\binom{n}{p} = \frac{n!}{p!(n-p)!}$$

Proposition Binomial Theorem

For any non-negative integer and any complex numbers $a$ and $b$, we have$$\begin{aligned}(a+b)^n&=\sum_{k=0}^{n} \binom{n}{k}a^{\,k}b^{\,n-k}= \binom{n}{0}a^0 b^n + \binom{n}{1}a^1b^{n-1} + \binom{n}{2}a^2b^{n-1} + \dotsb + \binom{n}{n}a^n b^0\\ &= \sum_{k=0}^{n} \binom{n}{k}a^{\,n-k}b^{\,k}= \binom{n}{0}a^n b^0 + \binom{n}{1}a^{n-1}b^1 + \binom{n}{2}a^{n-2}b^2 + \dotsb + \binom{n}{n}a^0 b^n\end{aligned}$$

Proof

Combinatorial Argument:
- Case $n=0$: By convention, $(a+b)^0 = 1$. The formula gives $\binom{0}{0}a^0b^0 = 1 \times 1 \times 1 = 1$. Thus, the theorem holds.
- Case $n > 0$: Consider the product$$(a+b)^n = \underbrace{(a+b)(a+b)\dotsm(a+b)}_{\text{$n$ factors}}.$$To obtain a term in the expansion, we choose either $a$ or $b$ from each factor and multiply the $n$ choices together.
  A term of the form $a^{n-k}b^k$ appears whenever we choose $b$ from exactly $k$ of the $n$ brackets (and $a$ from the remaining $n-k$ brackets).
  The number of such choices is precisely $\binom{n}{k}$, because we must choose which $k$ positions will contribute a factor $b$.
  Therefore the coefficient of $a^{n-k}b^k$ is $\binom{n}{k}$, and summing over all $k$ from $0$ to $n$ gives$$(a+b)^n = \sum_{k=0}^{n} \binom{n}{k}a^{\,n-k}b^{\,k}.$$
Summing all possible terms from $k=0$ to $k=n$ gives the result.
Proof by Induction:
Let $P(n)$ be the property: $(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$.
- Base case: For $n=0$, $(a+b)^0 = 1$.
  The formula gives $\sum_{k=0}^{0} \binom{0}{k} a^{0-k} b^k = \binom{0}{0} a^0 b^0 = 1$.
  Thus $P(0)$ is true.
- Inductive step: Assume $P(n)$ is true for some $n \ge 0$. Let's compute $(a+b)^{n+1}$: $$(a+b)^{n+1} = (a+b) \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k = \sum_{k=0}^{n} \binom{n}{k} a^{n-k+1} b^k + \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^{k+1}$$ By shifting the index of the second sum (let $j = k+1$), we combine the terms: $$(a+b)^{n+1} = \binom{n}{0}a^{n+1} + \sum_{k=1}^{n} \left[ \binom{n}{k} + \binom{n}{k-1} \right] a^{n-k+1} b^k + \binom{n}{n}b^{n+1}$$ Using Pascal's Identity, $\binom{n}{k} + \binom{n}{k-1} = \binom{n+1}{k}$, and noting that $\binom{n}{0}=\binom{n+1}{0}$ and $\binom{n}{n}=\binom{n+1}{n+1}$: $$(a+b)^{n+1} = \sum_{k=0}^{n+1} \binom{n+1}{k} a^{n+1-k} b^k.$$
Thus $P(n+1)$ is true, and the theorem is proven for all $n \in \mathbb{N}$.

Factorization of $a^n - b^n$

In addition to the expansion of $(a+b)^n$, it is essential to know how to factor the difference of two $n$-th powers: $a^n-b^n$.
This identity is a generalization of the well-known formula $a^2-b^2 = (a-b)(a+b)$.

Proposition General Factorization Formula

For any positive integer $n$ and any complex numbers $a$ and $b$:$$\begin{aligned} a^n - b^n &= (a - b) \sum_{k=0}^{n-1} a^{n-1-k} b^k\\ &= (a - b)(a^{n-1} + a^{n-2}b + a^{n-3}b^2 + \dots + b^{n-1})\\ \end{aligned}$$

Proof

Let us expand the product $(a-b) \sum_{k=0}^{n-1} a^{n-1-k} b^k$:$$\begin{aligned}(a-b) \sum_{k=0}^{n-1} a^{n-1-k} b^k &= a \sum_{k=0}^{n-1} a^{n-1-k} b^k - b \sum_{k=0}^{n-1} a^{n-1-k} b^k \\ &= \sum_{k=0}^{n-1} a^{n-k} b^k - \sum_{k=0}^{n-1} a^{n-1-k} b^{k+1} \\ &= \sum_{k=0}^{n-1} a^{n-k} b^k - \sum_{j=1}^{n} a^{n-j} b^j \quad (\text{change of index by setting } j=k+1) \\ &= \left( a^n + \sum_{k=1}^{n-1} a^{n-k} b^k \right) - \left( \sum_{j=1}^{n-1} a^{n-j} b^j + b^n \right) \quad \left( \text{extracting the first term of } \textstyle \sum_k \text{ and the last term of } \textstyle \sum_j \right) \\ &= a^n - b^n \quad (\text{the two sums are identical and cancel each other out})\end{aligned}$$Note: Instead of the formal index translation above, we can use an expansion method after the second step. By writing out the terms of both sums explicitly, we observe a telescoping sum:$$\begin{aligned}\text{Sum 1: } & a^n + a^{n-1}b + a^{n-2}b^2 + \dots + ab^{n-1} \\ \text{Sum 2: } & \phantom{a^n + } a^{n-1}b + a^{n-2}b^2 + \dots + ab^{n-1} + b^n\end{aligned}$$When we subtract the two, all the intermediate terms cancel out, leaving only:$$ a^n - b^n $$

Example

Factorize $z^3 - 8$ for $z \in \mathbb{C}$.

Answer

Recognize that $8 = 2^3$. Using the formula with $a=z$ and $b=2$:$$\begin{aligned}z^3 - 2^3 &= (z - 2)(z^2 + z^1 \times 2^1 + 2^2) \\ &= \mathbf{(z - 2)(z^2 + 2z + 4)}\end{aligned}$$

Binomial Theorem

Binomial Expansion for \(n\equal 2\) and \(n\equal 3\)

Pascal's Triangle

Binomial Theorem

Factorization of \(a^n - b^n\)