$C_n^m=C_{n-1}^{m-1}+C_{n-1}^m$
by Juanzhang @ 2018-11-23 23:30:43
@[NaCly_Fish](/space/show?uid=115864) 抱歉那个确实错了~~（不过很明显qaq~~
by Juanzhang @ 2018-11-23 23:31:27
@[小光](/space/show?uid=73934) QAQ我试试吧
by NaCly_Fish @ 2018-11-23 23:34:25
$\displaystyle\sum_{i=0}^{k}{C_{n+1}^i}=\displaystyle\sum_{i=0}^{k}{C_{n}^i}+\displaystyle\sum_{i=0}^{k-1}{C_{n}^i}$
$\displaystyle\sum_{i=0}^{k}{C_{n-1}^i}=\frac{(\displaystyle\sum_{i=0}^{k}{C_{n}^i})+C_n^{i-1}}{2}$
by Juanzhang @ 2018-11-23 23:35:28
@[NaCly_Fish](/space/show?uid=115864) 坐标？
by Juanzhang @ 2018-11-23 23:40:46