中山组合数学推了我一晚上

求

\( \begin{aligned} \cfrac{1}{2^{n+m}}\sum^{n}_{a=0}\sum^{m}_{b=0}\sum^{a+b}_{i=0}2i\dbinom{a}{i}\dbinom{b}{i} \end{aligned} \)

我们提取出 \(2^{-(n+m-1)}\)，根据 oi-wiki 二项式推论第十条则有

\( \begin{aligned} \sum^{n}_{a=0}\sum^{m}_{b=0}\sum^{a+b}_{i=0}i\dbinom{a}{i}\dbinom{b}{i} &=\sum^{n}_{a=0}\sum^{m}_{b=0}\sum^{\min(a,b)}_{i=1}i\dbinom{a}{i}\dbinom{b}{i}\\ &=\sum^{\min(n,m)}_{i=1}i \left( \sum^{n}_{a=0}\dbinom{a}{i}\right)\left(\sum^{m}_{b=0}\dbinom{b}{i}\right) \\ &=\sum^{n}_{i=1}i \dbinom{n+1}{i+1}\dbinom{m+1}{i+1} \end{aligned} \)

我们令 \(j=i+1,i=j-1\)，则有

\( \begin{aligned} \sum^{n}_{i=1}i \dbinom{n+1}{i+1}\dbinom{m+1}{i+1} &=\sum^{n+1}_{j=2}(j-1) \dbinom{n+1}{j}\dbinom{m+1}{j}\\ &=\sum^{n+1}_{j=0}(j-1) \dbinom{n+1}{j}\dbinom{m+1}{j}+1\\ &=\sum^{n+1}_{j=1}j \dbinom{n+1}{j}\dbinom{m+1}{j}-\sum^{n+1}_{j=0}\dbinom{n+1}{j}\dbinom{m+1}{j}+1\\ \end{aligned} \)

根据 oi-wiki 范德蒙德卷积第四条，和 oi-wiki 二项式推论第三条，

\( \begin{aligned} \sum^{n+1}_{j=0}\dbinom{n+1}{j}\dbinom{m+1}{j} &=\dbinom{n+m+2}{n+1}\\ &=\dbinom{n+m+1}{n+1}+\dbinom{n+m+1}{n} \end{aligned} \)

根据 oi-wiki 二项式推论第二条，我们有

\( \begin{aligned} j \dbinom{n+1}{j}=(n+1) \dbinom{n}{j-1} \end{aligned} \)

令 \(k=j-1,j=k+1\)，和 oi-wiki 范德蒙德卷积第四条

\( \begin{aligned} \sum^{n+1}_{j=1}j \dbinom{n+1}{j}\dbinom{m+1}{j} &=(n+1)\sum^{n+1}_{j=1}\dbinom{n}{j-1}\dbinom{m+1}{j}\\ &=(n+1)\sum^{n}_{k=0}\dbinom{n}{k}\dbinom{m+1}{k+1}\\ &=(n+1)\dbinom{n+m+1}{n+1} \end{aligned} \)

则

\( \begin{aligned} Ans&= (n+1)\dbinom{n+m+1}{n+1}-\left(\dbinom{n+m+1}{n+1}+\dbinom{n+m+1}{n}\right)+1\\ &=n\dbinom{n+m+1}{n}-\dbinom{n+m+1}{n}+1 \end{aligned} \)