1/sn=1/n(n+2)=[(1/n)-1/(n+2)]/2 一般都是朝着这个方向发展为抵消项作准备
s1=（1-1/3）/2
s2=(1/2-1/4)/2
s3=(1/3-1/5)/2
1/S1+1/S2+1/S3+……+1/Sn=[(1-1/3)+(1/2-1/4)+(1/3-1/5)+(1/4-1/6)+.+1/(n-1）-1/（n+1)+1/n-1/(n+2)]/2注意观察抵消的项
=[1+1/2-1/（n+1)-1/(n+2)]/2
=(3n^2+5n)/4(n+1)(n+2)
=n(3n+5)/4(n+1)(n+2)