Question

高中数学前n项和？

如图，这个式子怎么求出来的，求下过程，谢谢啦

Answer 1

n(n+1)(n+2)=[-(n-1)n(n+1)n+2) +n(n+1)(n+2)(n+3)]/4
就这样每项裂为两项差,抵消后剩下最后一项.

Answer 2

解，an=n^3-n
则sn=(1^3+2^3+，，+n^3)-(1+2+，，，+n)
=1/4(n+1)^2n^2-2/4(n+1)n
=1/4(n+1)n(n^2+n-2)
=1/4(n+1)n(n-1)(n+2)