摘要: 令\[S_i=\sum_{k=1}^n k^i m^k\]我们有\[\begin{eqnarray*}(m-1)S_i & = & mS_i - S_i \\& = & \sum_{k=1}^n k^i m^{k+1} - \sum_{k=1}^n k^i m^k \\& = & \sum_{k=2... 阅读全文
posted @ 2014-05-14 08:20 zhuohan123 阅读(1425) 评论(2) 推荐(1)