已知数列{an}的前n项和和an的关系为2Sn²=2anSn-an(n>1),a1=2求数列{1/Sn}是公差为2的等差数列

问题描述:

已知数列{an}的前n项和和an的关系为2Sn²=2anSn-an(n>1),a1=2求数列{1/Sn}是公差为2的等差数列

由2Sn²=2anSn—an 可得2Sn(Sn—an)= —an (n>1)又Sn—an = s(n-1) ;an = Sn — s(n-1) (n>1)故有 2 Sn•s(n-1) = —[Sn — s(n-1)]= s(n-1) — Sn (n>1)上式两边同时除以Sn•sn-1可得 1/ Sn...