求数列1/[n*(n+2)]前n项和

1/[n*(n+2)]=[1/n-1/(n+2)]/2
数列1/[n*(n+2)]前n项和
Sn=[1/1-1/3+1/2-1/4+1/3-1/5+1/4-1/6+...+1/(n-3)-1/(n-1)+1/(n-2)-1/n+1/(n-1)-1/(n+1)+1/n-1/(n+2)]/2
=[1/1+1/2-1/(n+1)-1/(n+2)]/2
=3/4-(2n+3)/[2(n+1)(n+2)]

没分！也告诉你吧

这种题目简单呀，一看就是要拆的

1/[n*(n+2)]=(1/2)*[(1/n)-1/(n+2)]

前N项和：
(1/2)*[(1 - 1/3)+(1/2 - 1/4)+(1/3 - 1/5)+(1/4 - 1/6)+...+(1/n - 1/(n+2))]
=(1/2)*[3/2 -(2n-3)/(n+1)(n+2)]

OK了自己再化简一下去验证绝对是正确的！！！