多項定理
= ( x 1 + ⋯ + x m ) ( x 1 + ⋯ + x m ) n {\displaystyle =(x_{1}+\cdots +x_{m})(x_{1}+\cdots +x_{m})^{n}}
= ( x 1 + ⋯ + x m ) ∑ k 1 + ⋯ + k m = n ( n k 1 , ⋯ , k m ) x 1 k 1 ⋯ x m k m {\displaystyle =(x_{1}+\cdots +x_{m})\textstyle \sum \limits _{k_{1}+\cdots +k_{m}=n}{\dbinom {n}{k_{1},\cdots ,k_{m}}}{x_{1}}^{k_{1}}\cdots {x_{m}}^{k_{m}}}
= x 1 ∑ k 1 + ⋯ + k m = n ( n k 1 , ⋯ , k m ) x 1 k 1 ⋯ x m k m + x m ∑ k 1 + ⋯ + k m = n ( n k 1 , ⋯ , k m ) x 1 k 1 ⋯ x m k m {\displaystyle =x_{1}\textstyle \sum \limits _{k_{1}+\cdots +k_{m}=n}{\dbinom {n}{k_{1},\cdots ,k_{m}}}{x_{1}}^{k_{1}}\cdots {x_{m}}^{k_{m}}+x_{m}\textstyle \sum \limits _{k_{1}+\cdots +k_{m}=n}{\dbinom {n}{k_{1},\cdots ,k_{m}}}{x_{1}}^{k_{1}}\cdots {x_{m}}^{k_{m}}}
二項定理と同様に、指数 n についての数学的帰納法で証明できる。
n =1 のとき、
( x 1 + x 2 + ⋯ + x m ) 1 = 1 x 1 + 1 x 2 + ⋯ + 1 x m , {\displaystyle (x_{1}+x_{2}+\cdots +x_{m})^{1}=1x_{1}+1x_{2}+\cdots +1x_{m},}
1 ! 1 ! 0 ! ⋯ 0 ! = 1 {\displaystyle {\frac {1!}{1!\,0!\,\cdots 0!}}=1}
より成り立つ。
ある n について成り立つと仮定する。
( x 1 + ⋯ + x m ) n = ∑ k 1 + ⋯ + k m = n ( n k 1 , ⋯ , k m ) x 1 k 1 ⋯ x m k m {\displaystyle (x_{1}+\cdots +x_{m})^{n}=\textstyle \sum \limits _{k_{1}+\cdots +k_{m}=n}{\dbinom {n}{k_{1},\cdots ,k_{m}}}{x_{1}}^{k_{1}}\cdots {x_{m}}^{k_{m}}}
より、
= ( x 1 + ⋯ + x m ) n + 1 {\displaystyle {\hphantom {=}}\;(x_{1}+\cdots +x_{m})^{n+1}}