分类：教育/科学-学习帮助
问题：高二数列2已知数列{an}的首项a1=1,其前n项的和Sn与an 
已知数列{an}的首项a1=1,其前n项的和Sn与an之间满足an=2Sn²/(2Sn-1) (n≥2).
(1)求证:数列{1/Sn}为等差数列;
(2)求数列{an}的通项公式.
回答：2Sn²/(2Sn-1) =An =Sn -S(n-1)
==> 1/Sn -1/S(n-1) =2 ==> {1/Sn}为等差数列
S1=a1=1, 1/Sn =1 ==> 1/Sn =2n-1, Sn =1/(2n-1)
An=Sn-S(n-1)=1/(2n-1) -1/(2n-3)=-2/(2n-1)(2n-3),(n≥2)