Question

已知an=n(n+2) 求Sn（在线急等）

sn=1*3+2*4+3*5+......+n(n+2)
求前N项和Sn

Answer 1

an=（n+1）²-1
Sn=2²-1+3²-1+4²-1+...+（n+1）²-1
=[（n+1）（n+2）（2n+3）]/6-1-n

Answer 2

an=n^2+2n

Sn=(1^2+2^2+...+n^2)+2(1+2+...+n)
=n(n+1)(2n+1)/6+2*(1+n)n/2
=n(n+1)*(2n+1)/6+n(n+1)
=n(n+1)[(2n+1)/6+1]
=n(n+1)(2n+7)/6