已知数列{an}的前n项和为Sn,且满足an+2Sn*Sn-1=0,a1=1/2.求证：{1/Sn}是等差数列

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an+2Sn*Sn-1=0
其中an=Sn-Sn-1代入上式：
Sn-Sn-1+2Sn*Sn-1=0
a1＝1/2,故Sn和Sn-1≠0,上式两边同除以Sn*Sn-1得：
1/Sn-1-1/Sn+2＝0
即：1/Sn-1/Sn-1＝2
{1/Sn}为等差数列,公差为2,首项1/S1=1/a1＝2
1/Sn=2+2(n-1)=2n
Sn=1/2n

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