数列{an}首项a1=3通项an与前n项的Sn之间满足2an=SnS(n-1) (n>=2)

（1） an = Sn - S(n-1)
2(Sn-S(n-1)) = SnS(n-1)
Sn = 2S(n-1) / (2-S(n-1))
1/Sn = 1/S(n-1) - 1/2
所以 {1/Sn}为 首项是1/3,公差是 -1/2 的等差数列
（2）1/Sn 的通项公式是 1/Sn = 1/3 + (-1/2)*(n-1) = -1/2 n + 5/6
Sn = 6/(5-3n)
an = 1/2 ( 6/(5-3n) * 6/(8-3n)),n>=2