已知数列{a
n}的前n项和S
n满足a
n+2S
nS
n-1=0 (n≥2),a
1=
,求a
n= ___ .
人气:237 ℃ 时间:2019-09-09 18:26:15
解答
: ∵an+2snsn-1=0(n≥2),∴sn-sn-1+2snsn-1=0.两边除以2snsn-1,并移向得出1Sn-1Sn-1=2(n≥2),∴{1Sn}是等差数列,公差d=2,1S1=1a1=2.∴1Sn=2+2(n-1)=2n,故Sn=12n.∴当n≥2时,an=Sn-Sn-1=12n-12(n-1)=-12...
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