已知12+22+32+…+n2=1/6n(n+1)(2n+1),则数列1×2,2×3,3×4,…,n(n+1)的前n项和为: _

问题描述:

已知12+22+32+…+n2=

1
6
n(n+1)(2n+1),则数列1×2,2×3,3×4,…,n(n+1)的前n项和为: ___

数列1×2,2×3,3×4,…,n(n+1)的通项为:an=n(n+1)=n2+n.
所以:Sn=a1+a2+…+an=(12+22+…+n2)+(1+2+…+n)
=

1
6
n(n+1)(2n+1)+
1
2
n(n+1)=
n(n+1)(n+2)
3

故答案为
n(n+1)(n+2)
3