Question

1*2+2*3+……+n(n+1)(N+2)=

Answer 1

Sn=1×2+2×3+3×4+……+n(n+1)=1²+1+2²+2+3²+3+……+n²+n
=(1+2+3+……+n)+(1²+2²+3²+……+n²)
=n(n+1)/2+(1²+2²+3²+……+n²)
关键求1²+2²+3²+……+n²如下 1²+2²+3²+……+n²=n(n+1)(2n+1)/6
证:（利用恒等式(n+1)³=n³+3n²+3n+1）：
(n+1)³-n³=3n²+3n+1,
n³-(n-1)³=3(n-1)²+3(n-1)+1
..............................
3³-2³=3×2²+3×2+1
2³-1³=3×1²+3×1+1.
把这n个等式两端分别相加，得：
(n+1)³-1=3(1²+2²+3²+……+n²)+3(1+2+3+...+n)+n,
由于1+2+3+...+n=n(n+1)/2,
代人上式得：
n³+3n²+3n=3(1²+2²+3²+……+n²)+3(n+1)n/2+n
整理后得：
1²+2²+3²+……+n²=n(n+1)(2n+1)/6
所以Sn=n(n+1)/2+(1²+2²+3²+……+n²)=n(n+1)/2+n(n+1)(2n+1)/6

=n(n²+3n+2)/3