\[\sum_{i}\left\langle\begin{matrix}n\\i\end{matrix}\right\rangle \binom{i+k}{n}=k^n \]
通俗的证明(具体数学的习题 6.15)是使用归纳法。
我们也可以对后面几个式子用二项式反演证明,而有如下过程:
\[\begin{aligned} z^n&=\sum_{i=0}^n{n\brace i}z^\underline i\\ &=\sum_{i=0}^n\binom {z}i{n\brace i}i!\\ &=\sum_{i=0}^n\binom{z}i\sum_{k=n-i}^n \left\langle\begin{matrix}n\\k\end{matrix}\right \rangle\binom k {n-i}\\ &=\sum_{k=0}^n\left \langle\begin{matrix}n\\k\end{matrix}\right \rangle\sum_{i=n-k}^n\binom {z}i\binom k {n-i}\\ &=\sum_{k=0}^n\left \langle\begin{matrix}n\\k\end{matrix}\right \rangle\binom {z+k}n \end{aligned}\]我们也可以组合意义证明。考虑在 \(n\) 阶排列插入 \(k\) 个
标签:begin,4.12,end,matrix,sum,恒等式,right,binom,Worpitzky From: https://www.cnblogs.com/british-union/p/18130860/Wor