摘要: 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 阅读全文
posted @ 2021-05-23 20:36 la-la-wanf 阅读(742) 评论(0) 推荐(0) 编辑