已知x1=1/3 xn+1=xn2+xn-1/4求证 数列lg(xn+1/2)是等比数列
人气：245 ℃ 时间：2020-05-06 13:52:42
解答
∵ x(n+1)=x²n+xn-1/4
∴x(n+1)+1/2=x²n+xn+1/4=(xn+1/2)²
两边取对数：
lg[x(n+1)+1/2]=lg(xn+1/2)²=2lg(xn+1/2)
∴lg[(x(n+1)+1/2]/lg(xn+1/2)=2 是常数
即 数列lg(xn+1/2)是等比数列.
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