# 多項式定理

${\displaystyle (x_{1}+x_{2}+\cdots +x_{t})^{n}=\sum {\frac {n!}{n_{1}!n_{2}!\cdots n_{t}!}}x_{1}^{n_{1}}x_{2}^{n_{2}}\cdots x_{t}^{n_{t}}}$

${\displaystyle n_{1},n_{2},n_{3}\cdots n_{t}}$ 是指一切滿足上述條件的非負數組合。 由隔板法可知該多項式展開共有 ${\displaystyle {\frac {(n+t-1)!}{n!(t-1)!}}}$ 項。

## 證明

### 數學歸納法

${\displaystyle \left(x_{1}+x_{2}+\cdots +x_{t}\right)^{n}}$
${\displaystyle =}$  ${\displaystyle ((x_{1}+x_{2}+\cdots +x_{t-1})+x_{t})^{n}}$
${\displaystyle =}$  ${\displaystyle \sum _{n_{t}=0}^{n}{\frac {n!}{n_{t}!\left(n-n_{t}\right)!}}\left(x_{1}+x_{2}+\cdots +x_{t-1}\right)^{n-n_{t}}x_{t}^{n_{t}}}$
${\displaystyle =}$  ${\displaystyle \sum _{n_{t}=0}^{n}{\frac {n!}{n_{t}!\left(n-n_{t}\right)!}}\sum _{n_{1}+n_{2}+\cdots +n_{t-1}=n-n_{t}}{\frac {\left(n-n_{t}\right)!}{n_{1}!\cdots n_{t-1}!}}x_{1}^{n_{1}}\cdots x_{t-1}^{n_{t-1}}x_{t}^{n_{t}}}$
${\displaystyle =}$  ${\displaystyle \sum _{n_{1}+n_{2}+\cdots +n_{t}=n}{\frac {n!}{n_{1}!\cdots n_{t}!}}x_{1}^{n_{1}}\cdots x_{t}^{n_{t}}}$

### 組合法

${\displaystyle n_{1}+n_{2}+\cdots +n_{t}=n}$ 中選${\displaystyle n_{i}}$ ${\displaystyle x_{i}}$

${\displaystyle \displaystyle {\binom {n}{n_{1}}}{\binom {n-n_{1}}{n_{2}}}{\binom {n-n_{1}-n_{2}}{n_{3}}}\cdots {\binom {n-n_{1}-n_{2}-\cdots -n_{t-1}}{n_{t}}}}$

${\displaystyle ={\frac {n!(n-n_{1})!(n-n_{1}-n_{2})!\cdots (n-n_{1}-n_{2}-\cdots -n_{t-1})!}{n_{1}!(n-n_{1})!n_{2}!(n-n_{1}-n_{2})!n_{3}!(n-n_{1}-n_{2}-n_{3})!\cdots n_{t}!(n-n_{1}-n_{2}-\cdots -n_{t})!}}={\frac {n!}{n_{1}!n_{2}!n_{3}!\cdots n_{t}!}}}$ [1][2]

## 參考資料

1. ^ 陳景潤. 组合数学简介. 哈爾濱工業大學出版社. : 第81–83頁 [2015-09-20]. ISBN 9787560335643. （原始內容存檔於2017-04-13）.
2. ^ 伍啟期. 多项式定理的新证明及其展开. 佛山科學技術學院學報(自然科學版). 2012, (6) [2013-10-07]. （原始內容存檔於2017-04-13）.