已知函数f(x)=2x-1/2x,数列{an}满足f(log2an)=-2n.(1)求数列{an}的通项公式;(2)证明数列{an}是递减数列.
问题描述:
已知函数f(x)=2x-
,数列{an}满足f(log2an)=-2n.1 2x
(1)求数列{an}的通项公式;
(2)证明数列{an}是递减数列.
答
(1) ∵f(x)=2x-12x,f(log2an)=-2n,∴2log2an-2-log2an=-2n,an-1an=-2n,∴an2+2nan-1=0,解得an=-n±n2+1,∵an>0,∴an=n2+1-n,n∈N*;(2)证明:an+1an=(n+1)2+1-(n+1)n2+1-n=n2+1+n(n+1)2+1+(n+1)...