Question

已知数列{an}满足a1+2a2+3a3+…+nan=n(n+1)(n+2),则它的前n项和Sn=?

Sn=(3n^2+9n)/2_______
在线等，求详解。

Answer 1

a1+2a2+3a3+…+nan=n(n+1)(n+2)
a1+2a2+3a3+…+(n-1)a(n-1)=(n-1)(n)(n+1)
两式相减，得
nan=3n(n+1)
an=3n+3

Sn=(a1+an)*n/2=(3n+9)n/2=(3n^2+9n)/2

Answer 2

a1+2a2+3a3+…+nan=n(n+1)(n+2), (1)
a1+2a2+3a3+…+(n-1)an-1=(n-1)(n+1)n (2)
(1)-(2)得：nan=n(n+1)(n+2-n+1)
an=3(n+1)
Sn=3n[2+n+1]/2=(3n^2+9n)/2

追问

谢谢