已知数列{an}a1=1 n>=2 Sn^2=an(Sn-1/2)证明是{1/Sn}等差数列

Sn^2=an[Sn-1/2]
因为an=Sn-S(n-1)代入上式得到
Sn^2=(Sn-S(n-1))[Sn-1/2]
整理S(n-1)-Sn=2SnS(n-1)
左右同时除以SnS(n-1)得到
1/Sn-1/S(n-1)=2
所以{1/Sn}是等差数列,公差为2,首项为1/S1=1