Toolkit 20

Binomial Theorem

(a+b)^n=\binom{n}{0}a^n+\binom{n}{1}a^{n-1}b+\binom{n}{2}a^{n-2}b^2+\cdots+\binom{n}{n}b^n

Proof

Write

(a+b)^n = \underbrace{(a+b)(a+b)\cdots(a+b)}_{n\text{ factors}}.

To obtain a term $a^{n-k}b^k$ , we must choose $b$ from exactly $k$ of the $n$ factors and choose $a$ from the remaining $n-k$ factors.

There are

\binom{n}{k}

ways to choose the $k$ factors that contribute $b$ . Therefore, the coefficient of $a^{n-k}b^k$ is $\binom{n}{k}$ .

Hence,

(a+b)^n = \binom{n}{0}a^n+\binom{n}{1}a^{n-1}b+\binom{n}{2}a^{n-2}b^2+\cdots+\binom{n}{n}b^n. \quad\square