已知:Sn为数列bn的前几项和,且满足(2bn)/(bnSn-Sn^2)=1 ,b1=1.
问题描述:
已知:Sn为数列bn的前几项和,且满足(2bn)/(bnSn-Sn^2)=1 ,b1=1.
证明:(1)数列{1/Sn}成等差(2)求数列{bn}的通项公式
答
由(2bn)/(bnSn-Sn^2)=1
=〉 Sn^2-bnSn+2bn=0
将bn=Sn-S[n-1] (n-1是下标)(n>0)
Sn^2-(Sn-S[n-1])Sn+2(Sn-S[n-1])=0
=>S[n-1])Sn+2(Sn-S[n-1])=0
=>1/Sn-1/S[n-1]=1/2
即数列{1/Sn}成等差数列,公差为1/2
又b1=1 所以 1/S1=1
所以数列{1/Sn}通式为1+(n-1)/2=(n+1)/2
所以 Sn=2/(n+1)
=> bn=Sn-S[n-1]=2/(n+1)-2/n=-2/[n(n+1)]