已知数列{an}的前n项之和Sn与an之间满足2Sn^2=2anSn-an (n>=2),且a1=2

题目

已知数列{an}的前n项之和Sn与an之间满足2Sn^2=2anSn-an (n>=2),且a1=2
求证：1.数列{1/Sn}是以2为公差的等差数列
2.求Sn和an

答案

1.证：n≥2时,2Sn²=2anSn-an=2[Sn-S(n-1)]Sn-[Sn-S(n-1)]整理,得S(n-1)-Sn=2SnS(n-1)等式两边同除以SnS(n-1)1/Sn -1/S(n-1)=2,为定值.1/S1=1/a1=1/2,数列{1/Sn}是以1/2为首项,2为公差的等差数列.2.1/Sn=(1/2)+2...