1^2+2^2+3^2+……+n^2=n(n+1)(2n+1)/6
1^3+2^3+3^3+...+n^3=(1+2+3+...+n)^2 =[n(n+1)/2]^2
1^4+2^4+3^4+……+n^4=n(n+1)(2n+1)(3n2+3n-1)/30.
1^5+2^5+3^5+……+n^5=n^2(n+1)^2(2n2+2n-1)/12.
几个i的幂的累加公式1^2+2^2+3^2 2~5
原文：https://www.cnblogs.com/la-la-wanf/p/14802090.html
