多項式定理

多項式定理為二項式定理的推廣。${\displaystyle t=2}$ 時為二項式定理。

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

其中 ${\displaystyle n_{1}+n_{2}+\cdots +n_{t}=n}$、${\displaystyle 0\leq n_{i}\leq n}$

${\displaystyle n_{1},n_{2},n_{3}\cdots n_{t}}$ 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

對元數t做歸納： 當t=2時，原式為二項式定理，成立。 假設對t-1元成立，則：

${\displaystyle \left(x_{1}+x_{2}+\cdots +x_{t}\right)^{n}}$

${\displaystyle =}$ ${\displaystyle ((x_{1}+x_{2}+\cdots +x_{t-1})+x_{t})^{n}}$

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

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

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

證畢.

組合法

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

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

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

參見

• 二項式定理
• 多項式