数列（n^2）*（2n-1）求和

题目

数列（n^2）*（2n-1）求和

答案

an
=n^2(2n-1)
=2n(n+1)(n+2) - 7n(n+1) +3n
=(1/2)[n(n+1)(n+2)(n+3) -(n-1)n(n+1)(n+2)]- (7/3)[n(n+1)(n+2) -(n-1)n(n+1)] + (3/2)[n(n+1) -(n-1)n]
Sn = a1+a2+...+an
=(1/2)n(n+1)(n+2)(n+3) - (7/3)n(n+1)(n+2) + (3/2)n(n+1)
= (1/6)n(n+1) [ 3(n+2)(n+3) - 14(n+2)+ 9]
= (1/6)n(n+1) ( 3n^2+15n+18 -14n-28+9]
=(1/6)n(n+1)(3n^2+n-1)