S(n,m)=∑{i=1,n} i^m
扰动法
S(n,m+1) = ∑ i(m+1) S(n+1,m+1) = ∑ i(m+1)
S(n+1,m+1)-S(n,m+1) = (n+1)(m+1)
S(n+1,m+1)-S(n,m+1) = 1+∑ (i+1)(m+1)-i(m+1) (错位相减)
(n+1)(m+1) = 1+∑ (i+1)(m+1)-i(m+1) (第一个循环为1到n)
= 1+∑ -i(m+1) +∑ C(m+1,j) *ij (二项式展开)
= 1+∑ ∑ C(m+1,j) *ij (这里第二个循环的上界变成m)
= 1+∑ C(m+1,j) ∑ ij (交换求和顺序)
= 1+∑ C(m+1,j) S(
(笔者过菜,不会敲数学公式,请读者自行想象sigma的上下界)
标签:相减,自然,过菜,ij,幂数,sigma,循环 From: https://www.cnblogs.com/zhuzc/p/18097237