Question

已知数列{an}的通项公式为an=1/n(n+2）求前n项的和

Sn=1/(1×3)+1/(2×4)+1/(3×5)+1/(4×6)+···1/[(n-1)(n+1)]+1/[n(n+2)]
=(1/2)[(1-1/3)+(1/2-1/4)+(1/3-1/5)···(1/n-1/(n+2))]
=(1/2)[1+1/2-1/(n+1)-1/(n+2)]
第三行的-1/(n+1)-1/(n+2)是怎么来的

Answer 1

在最后出现+[(1/(n-1)-1/(n+1)]+[1/n-1/(n+2)],其中1/(n-1)与1/n都与前面的项抵消了，就剩下最后2项。实际上，第二行的中括号内化简后就是最前2项和最后2项。

Answer 2

Sn=1/(1×3)+1/(2×4)+1/(3×5)+1/(4×6)+···1/[(n-1)(n+1)]+1/[n(n+2)]
=(1/2)[(1-1/3)+(1/2-1/4)+(1/3-1/5)···(1/n-1/(n+2))]
=(1/2)[(1-1/3)+(1/2-1/4)+(1/3-1/5)+...+(1/(n-2)-1/n)+(1/(n-1)-1/(n+1))+(1/n-1/(n+2))]
=(1/2)[1+1/2-1/(n+1)-1/(n+2)]
可以看出来，都互相减掉了，剩下这两项。