An*An+1=(1/2)n次方,Bn=A2n,求证{Bn}为等比数列
问题描述:
An*An+1=(1/2)n次方,Bn=A2n,求证{Bn}为等比数列
A1=1
答
有已知 A1=1,得:A2=1/2An*A(n+1)=(1/2)^n,A(n-1)*An=(1/2)^(n-1)两式相除得到:A(n+1)/A(n-1)=1/2有上式有A4/A2=A6/A4=.=1/2即{A2n}是公比为1/2的等比数列,首项A2=1/2,所以A2n=(1/2)^n ,Bn=A2n因为{A2n}是等比数列,...