2.已知数列{an}的前n项和为Sn,且满足a1=1/2,an+?

2.已知数列{an}的前n项和为Sn,且满足a1=1/2,an+?: 已知数列{an}的前n项和为Sn,且满足a1=1/2,an+2SnSn-1=0(n>=2) 判断{1/Sn}是否为等差数列？并证明，求Sn和an; n≥2时,An=Sn-S(n-1), ∴ Sn-S(n-1)+2SnS(n-1),两边除以SnS(n-1)=0,得(1/Sn)-1/S(n-1)=-2, ∴ {1/Sn}是等差数列.又A1=1/2, 1/S1=2, ∴ 1/Sn=2+(n-1)(-2)=-2n+4, Sn=1/(4-2n). n≥2时,An=-2SnS(n-1)=-1/[2(2-n)(1-n)]. ∴ Sn=1/(4-2n); An=1/2(n=1), An=-1/[2(2-n)(1-n)](n≥2).