> 数学 >
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.
人气:414 ℃ 时间:2020-06-21 14:40:06
解答
(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)...
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