1n+2(n-1)+3(n-2)+…+n1的前n项和_数学解答

1n+2(n-1)+3(n-2)+…+n1的前n项和

人气：273 ℃　时间：2020-06-15 20:22:10

解答

通项ak=k(n+1-k)=k(n+1)-k^2
Sk=(n+1)(1+2+……+n)-(1^2+2^2+……+n^2)
=n(n+1)^2/2-n(n+1)(2n+1)/6
=n(n+1)(n+2)/6