求数列1/（n^2+2n）前n项和

1/（n^2+2n）=[1/n - 1/(n+2)]/2

前n项和=(1-1/3)/2+(1/2-1/4)/2+...+[1/n-1/(n+2)]/2
裂项相消
=[1+1/2-1/(n+1)-1/(n+2)]/2
=3/4-(2n+3)/[2(n+1)(n+2)]

由1/(n^2+2n)=(1/n-1/(n+2))/2,故
1/(1^2+2*2)+1/(2^2+2*2)+1/(3^2+3*2)+..+1/(n^2+2n)
=(1/1-1/3+1/2-1/4+1/3-1/5+...+1/n-1/(n+2))/2
=(1+1/2-1/(n+1)-1/(n+2))/2
=(3/4-(2n+3)/(2(n+1)(n+2)))=(3n^2+5n)/(2(n+1)(n+2))

1/（n^2+2n)=[1/n-1/(n+2)]/2
Sn=[1/1-1/3]/2+[1/2-1/4]/2+[1/3-1/5]/2+...+[1/n-1/(n+2)]/2
=[1+1/2-1/(n+1)-1/(n+2)]/2
=[3/2-1/(n+1)-1/(n+2)]/2

首先 1/(n^2+2n)=1/2n-1/2(n+2)
则1/(1^2+2)+1/(2^2+2x2)+……+1/（n^2+2n）
=1/2(（1/1-1/3+1/2-1/4+1/3-1/15+……+1/n-1/(n+2)）
=1/2((1/1+1/2-1/(n+1)-1/(n+2))
=3/4-1/2(n+1)(n+2)

haoliao

1/（n^2+2n）等于1/(n*（n+2）)然后裂项，1/2*（1/n-1/（n+2））然后把n=1到n=n的项都列出来，把1/2提出来，相邻项都可以消掉了。会了吧？