神秘东西

交换上下指标

对于正整数 $n,m$：

$$(\genfrac{}{}{0ex}{}{n}{m})=(\genfrac{}{}{0ex}{}{n}{n-m})=(-1{)}^{n-m}(\genfrac{}{}{0ex}{}{-m-1}{n-m}).$$

试看看！

$$\begin{array}{rl}& \sum _{-q\le k\le l}(\genfrac{}{}{0ex}{}{q+k}{m})(\genfrac{}{}{0ex}{}{l-k}{n})\\ & =\sum _{0\le k\le q+l}(\genfrac{}{}{0ex}{}{k}{m})(\genfrac{}{}{0ex}{}{q+l-k}{n})\\ & =\sum _{m\le k\le q+l-n}(\genfrac{}{}{0ex}{}{k}{m})(\genfrac{}{}{0ex}{}{q+l-k}{n})\\ & =\sum _{0\le k\le q+l-n-m}(-1{)}^{k-m}(\genfrac{}{}{0ex}{}{-m-1}{k-m})(-1{)}^{q+l-k-n}(\genfrac{}{}{0ex}{}{-n-1}{q+l-k-n})\\ & =(-1{)}^{q+l-n-m}(\genfrac{}{}{0ex}{}{-m-n-2}{q+l-n-m})\\ & =(\genfrac{}{}{0ex}{}{q+l+1}{q+l-n-m})\\ & =(\genfrac{}{}{0ex}{}{q+l+1}{n+m+1}).\end{array}$$