一道等差数列题,
已知数列(an)的前n项和Sn,且满足an+2Sn*Sn-1=0(n>=2),a1=1/2,(1)求证(1\Sn)是等差数列
(2)求an
人气:306 ℃ 时间:2020-06-22 00:27:49
解答
an=Sn-Sn-1
an+2Sn*Sn-1=0
Sn-Sn-1+2Sn*Sn-1=0
除以Sn*Sn-1
1/Sn-1/Sn-1=2
(1\Sn)是等差数列
S1=a1=1/2
1/Sn=2n
Sn=1/2n
an=Sn-Sn-1=-1/(n-1)n,n≥2
n=1时,a1=1/2
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