范德蒙德卷积以及一些二项式系数的推导

1 : ${\sum }_{i=0}^k\left(\genfrac{}{}{0ex}{}{n}{i}\right)\left(\genfrac{}{}{0ex}{}{m}{k-i}\right)=\left(\genfrac{}{}{0ex}{}{n+m}{k}\right)$

从意义上理解即可，也就是从数量为$n$和$m$的两个堆中一共选择$k$个物品

这两个堆在实际意义上也可以不存在。

2 : ${\sum }_{i=1}^n\left(\genfrac{}{}{0ex}{}{n}{i}\right)\left(\genfrac{}{}{0ex}{}{n}{i-1}\right)=\left(\genfrac{}{}{0ex}{}{2n}{n-1}\right)$

证明 :

设$k=n-1$则由1得 :

${\sum }_{i=0}^{n-1}\left(\genfrac{}{}{0ex}{}{n}{i}\right)\left(\genfrac{}{}{0ex}{}{n}{n-1-i}\right)={\sum }_{i=0}^{n-1}\left(\genfrac{}{}{0ex}{}{n}{i}\right)\left(\genfrac{}{}{0ex}{}{n}{i+1}\right)={\sum }_{i=1}^n\left(\genfrac{}{}{0ex}{}{n}{i}\right)\left(\genfrac{}{}{0ex}{}{n}{i-1}\right)$

${\sum }_{i=0}^{n-1}\left(\genfrac{}{}{0ex}{}{n}{i}\right)\left(\genfrac{}{}{0ex}{}{n}{n-1-i}\right)=\left(\genfrac{}{}{0ex}{}{2n}{n-1}\right)\Rightarrow {\sum }_{i=1}^n\left(\genfrac{}{}{0ex}{}{n}{i}\right)\left(\genfrac{}{}{0ex}{}{n}{i-1}\right)=\left(\genfrac{}{}{0ex}{}{2n}{n-1}\right)$

证毕。

或者是

$\left(\genfrac{}{}{0ex}{}{2n}{n-1}\right)={\sum }_{i=0}^n\left(\genfrac{}{}{0ex}{}{n}{i}\right)\left(\genfrac{}{}{0ex}{}{n}{n-1-i}\right)={\sum }_{i=0}^n\left(\genfrac{}{}{0ex}{}{n}{i}\right)\left(\genfrac{}{}{0ex}{}{n}{i+1}\right)={\sum }_{i=1}^n\left(\genfrac{}{}{0ex}{}{n}{i}\right)\left(\genfrac{}{}{0ex}{}{n}{i-1}\right)$

其他的一些 :

${\sum }_{i=0}^n{\left(\genfrac{}{}{0ex}{}{n}{i}\right)}^2={\sum }_{i=0}^n\left(\genfrac{}{}{0ex}{}{n}{i}\right)\left(\genfrac{}{}{0ex}{}{n}{n-i}\right)=\left(\genfrac{}{}{0ex}{}{2n}{n}\right)$

${\sum }_{i=0}^m\left(\genfrac{}{}{0ex}{}{n}{i}\right)\left(\genfrac{}{}{0ex}{}{m}{i}\right)={\sum }_{i=0}^m\left(\genfrac{}{}{0ex}{}{n}{i}\right)\left(\genfrac{}{}{0ex}{}{m}{m-i}\right)=\left(\genfrac{}{}{0ex}{}{n+m}{m}\right)$

。

例题：

$[AHOI/HNOI2017]抛硬币$

其他的一些二项式系数推导的积累：

$$\sum _{k=1}^n{k}^2\left(\genfrac{}{}{0ex}{}{n}{k}\right)=n(n+1)2^{n-2}$$
过程：$k^2$转下降幂后暴力展开组合数即可
$$\sum _{k=1}^{n}k{\left(\genfrac{}{}{0ex}{}{n}{k}\right)}^2=n\left(\genfrac{}{}{0ex}{}{2n-1}{n-1}\right)$$
过程：
$$k\left(\genfrac{}{}{0ex}{}{n}{k}\right)=n\left(\genfrac{}{}{0ex}{}{n-1}{k-1}\right)=n\left(\genfrac{}{}{0ex}{}{n-1}{n-k}\right)$$
范德蒙德卷积即可。