이항정리

n이 양의 정수일 때

$$\begin{array}{rl}(a+b{)}^n& =\sum _{r=0}^n{}_n{C}_r{a}^{n-r}{b}^r=\sum _{r=0}^n\left(\begin{array}{c}n\\ r\end{array}\right)a^{n-r}{b}^r\\ & =\left(\begin{array}{c}n\\ 0\end{array}\right)a^n+\left(\begin{array}{c}n\\ 1\end{array}\right)a^{n-1}b+\cdots +\left(\begin{array}{c}n\\ r\end{array}\right)a^{n-r}{b}^r+\cdots +\left(\begin{array}{c}n\\ n\end{array}\right)b^n\end{array}$$

여기서 ${}_n{C}_r=\left(\begin{array}{c}n\\ r\end{array}\right)$은 n개에서 r개를 고르는 조합이다.

[증명] 이항계수의 항등식과 수학적 귀납법을 사용한다.
n=0 일 때 $(a+b{)}^0=1$
n에 대해 성립한다고 가정한다.
$\begin{array}{rl}(a+b{)}^{n+1}& =(a+b)(a+b{)}^n=(a+b)\sum _{r=0}^n\left(\begin{array}{c}n\\ r\end{array}\right)a^{n-r}{b}^r\\ & =a\sum _{r=0}^n\left(\begin{array}{c}n\\ r\end{array}\right)a^{n-r}{b}^r+b\sum _{r=0}^n\left(\begin{array}{c}n\\ r\end{array}\right)a^{n-r}{b}^r\\ & =\sum _{r=0}^n\left(\begin{array}{c}n\\ r\end{array}\right)a^{n-r+1}{b}^r+\sum _{r=1}^{n+1}\left(\begin{array}{c}n\\ r-1\end{array}\right)a^{n-r+1}{b}^r\\ & =a^{n+1}+b^{n+1}+\sum _{r=1}^n\{\left(\begin{array}{c}n\\ r\end{array}\right)+\left(\begin{array}{c}n\\ r-1\end{array}\right)\}{a}^{n-r+1}{b}^r\\ & =a^{n+1}+b^{n+1}+\sum _{r=1}^n\left(\begin{array}{c}n+1\\ r\end{array}\right)a^{n+1-r}{b}^r\\ & =\sum _{r=0}^{n+1}{a}^{n+1-r}{b}^r\end{array}$