多项式定理为二项式定理的推广。 t = 2 {\displaystyle t=2} 时为二项式定理。
( x 1 + x 2 + ⋯ + x t ) n = ∑ n ! n 1 ! n 2 ! ⋯ n t ! x 1 n 1 x 2 n 2 ⋯ x t n t {\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 n 1 + n 2 + ⋯ + n t = n {\displaystyle n_{1}+n_{2}+\cdots +n_{t}=n} 、 0 ≤ n i ≤ n {\displaystyle 0\leq n_{i}\leq n}
n 1 , n 2 , n 3 ⋯ n t {\displaystyle n_{1},n_{2},n_{3}\cdots n_{t}} 是指一切满足上述条件的非负数组合。由隔板法可知该多项式展开共有 ( n + t − 1 ) ! n ! ( t − 1 ) ! {\displaystyle {\frac {(n+t-1)!}{n!(t-1)!}}} 项。
对元数t做归纳:当t=2时,原式为二项式定理,成立。 假设对t-1元成立,则:
从 n 1 + n 2 + ⋯ + n t = n {\displaystyle n_{1}+n_{2}+\cdots +n_{t}=n} 中选 n i {\displaystyle n_{i}} 个 x i {\displaystyle x_{i}} :
( n n 1 ) ( n − n 1 n 2 ) ( n − n 1 − n 2 n 3 ) ⋯ ( n − n 1 − n 2 − ⋯ − n t − 1 n t ) {\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}}}}
= n ! ( n − n 1 ) ! ( n − n 1 − n 2 ) ! ⋯ ( n − n 1 − n 2 − ⋯ − n t − 1 ) ! n 1 ! ( n − n 1 ) ! n 2 ! ( n − n 1 − n 2 ) ! n 3 ! ( n − n 1 − n 2 − n 3 ) ! ⋯ n t ! ( n − n 1 − n 2 − ⋯ − n t ) ! = n ! n 1 ! n 2 ! n 3 ! ⋯ 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}!}}} 证毕.