已知数列{an}的前n项和Sn,且满足an+2SnSn-1=0.(n大于等于2)

an+2SnS(n-1)=0.(n大于等于2)
an=-2SnS(n-1)
an=Sn-S(n-1)
-2SnS(n-1)=Sn-S(n-1)
两边同除以-SnS(n-1),得
2=1/Sn-1/S(n-1)
所以,{1/Sn}是公差为2的等差数列,
S1=a1=1/2
1/S1=2
1/Sn=n(n+1)
Sn=1/n(n+1)
n=1时,a1=1/2
n>1时,有:
an=Sn-S(n-1)
=1/n(n+1)-1/(n-1)n
=-2/n(n^2-1)