一个小题目

$\displaystyle{\sum_{i=1}^n\sum_{j=1}^m\sum_{k=i}^n\sum_{l=j}^m\sum_{x=i}^k\sum_{y=j}^l}1$

$\Theta(1)$求上式
$\begin{aligned} \displaystyle{\sum_{i=1}^n\sum_{j=1}^m\sum_{k=i}^n\sum_{l=j}^m\sum_{x=i}^k\sum_{y=j}^l}1=&\displaystyle{\sum_{i=1}^n\sum_{j=1}^m\sum_{k=i}^n\sum_{l=j}^m}(k-i+1)(l-j+1)\\ =&\displaystyle{\sum_{i=1}^n\sum_{j=1}^m}\sum_{k=i}^n(k-i+1)\sum_{l=j}^m(l-j+1)\\ =&\displaystyle{\sum_{i=1}^n\sum_{j=1}^m}\frac{(n-i+1)(n-i+2)}{2}\frac{(m-j+1)(m-j+2)}{2}\\ =&\displaystyle{\sum_{i=1}^n}\frac{(n-i+1)(n-i+2)}{2}\sum_{j=1}^m\frac{(m-j+1)(m-j+2)}{2}\\ =&\displaystyle{\sum_{i=1}^n}\frac{n^{2}-2ni+3n+i^{2}-3i+2}{2}\sum_{j=1}^m\frac{m^{2}-2mi+3m+i^{2}-3i+2}{2}\\ =&\frac{n^3+3n^{2}+2n-\frac{2n^{2}(n+1)}{2}+\frac{n(n+1)(2n+1)}{6}-\frac{3n(n+1)}{2}}{2}\frac{m^3+3m^{2}+2m-\frac{2m^{2}(m+1)}{2}+\frac{m(m+1)(2m+1)}{6}-\frac{3m(m+1)}{2}}{2}\\ \end{aligned}$
成功了， 虽然我化的很麻烦