已知一个数列1/1X2,1/2X3,1/3X4,…1/n(n+1)…,则前n项的和Sn=?

Question

如题所述

Answer 1

此类题最常见的解法是拆项法来解
因为1/n(n+1)=1/n-1/(n+1)
所以
Sn=1/1X2+1/2X3+1/3X4+…+1/n(n+1）
=（1-1/2）+（1/2-1/3）+（1/3-1/4）+…+（1/n-1/(n+1)）
=1-1/(n+1)
=n/(n+1)
求1/1X3+1/3X5+1/5X7,…1/（2n-1)(2n+1)也是用拆项法来解
利用1/（2n-1)(2n+1)=(1/2)[1/（2n-1)-1/(2n+1)]

Answer 2

Sn=1/1X2+1/2X3+1/3X4+…+1/n(n+1）
=1-1/2+1/2-1/3+1/3-1/4+…+1/n-1/(n+1)
=1-1/(n+1)
=n/(n+1)

Answer 3

1/(1*2)+1/(2*3)+……+1/[n*(n+1)]
=1-1/2+1/2-1/3+……+1/n-1/(n+1)
=1-1/(n+1)

Answer 4

把每一项都拆成两项的差，再求就行