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=
n(n+1)(2n+1),则数列1×2,2×3,3×4,…,n(n+1)的前n项和为: ___ 1 6
答
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)
=
n(n+1)(2n+1)+1 6
n(n+1)=1 2
.n(n+1)(n+2) 3
故答案为
.n(n+1)(n+2) 3