数列{a
n}首项a
1=1,前n项和S
n与a
n之间满足
an=(n≥2)(1)求证:数列
{}是等差数列
(2)求数列{a
n}的通项公式.
人气:491 ℃ 时间:2019-10-24 04:52:05
解答
(1)∵当n≥2时,
an=Sn−Sn−1=,
整理得:S
n-1-S
n=2S
n⋅S
n-1,
由题意知S
n≠0,
∴
−=2,
即
{}是以
==1为首项,公差d=2的等差数列.
(2)∵
{}是以
==1为首项,公差d=2的等差数列.
∴
=1+2(n−1)=2n−1,
∴
Sn=,n∈N•.,
当n≥2时,
an=Sn−Sn−1=−=−,
当n=1时,a
1=S
1=1不满足a
n,
∴
an=.
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