数列{an}首项a1=1,前n项和Sn与an之间满足an=2Sn22Sn−1(n≥2) (1)求证:数列{1/Sn}是等差数列 (2)求数列{an}的通项公式.
问题描述:
数列{an}首项a1=1,前n项和Sn与an之间满足an=
(n≥2)2Sn2
2Sn−1
(1)求证:数列{
}是等差数列 1 Sn
(2)求数列{an}的通项公式.
答
(1)∵当n≥2时,an=Sn−Sn−1=2S2n2Sn−1,整理得:Sn-1-Sn=2Sn⋅Sn-1,由题意知Sn≠0,∴1Sn−1Sn−1=2,即{1Sn}是以1S1=1a1=1为首项,公差d=2的等差数列.(2)∵{1Sn}是以1S1=1a1=1为首项,公差d=2的等...