Sn=n^2+2n 求1/S1+1/S2+1/S3+……+1/Sn

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Sn=n^2+2n 求1/S1+1/S2+1/S3+……+1/Sn
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Sn=n^2+2n 求1/S1+1/S2+1/S3+……+1/Sn
Sn=n^2+2n 求1/S1+1/S2+1/S3+……+1/Sn

Sn=n^2+2n 求1/S1+1/S2+1/S3+……+1/Sn
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)

3/4
1/(n^2+2n)=1/2(1/n-1/(n+2))
约出来就只剩1/2（1+1/2-1/（n+2））=3/4

将1/Sn化简成1/2[1/n-1/（n+2）]
然后所求的式子就化成了1+1/2-1/(n+1）-1/（n+2）
结果为（3n*n+4n+3）/2(n+1)(n+2)

1/Sn=1/(n^2+2n)=(1/n-1/(n+2))/2
原式=(n+1)/2(n+2)

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