好吧问答库 > 已知正数数列{an}的前n项和为Sn,且4a-2sn=1,数列{bn}满足bn=2log2分之一an,n属于N.

已知正数数列{an}的前n项和为Sn,且4a-2sn=1,数列{bn}满足bn=2log2分之一an,n属于N.

2025年04月06日 23:16
有1个网友回答
网友（1）：

4an-2sn=1
4an+1-2sn+1=1
下式减上式得：2an+1=4an；an+1=2an
又a1=1/2；得an=2^(n-2)

bn=-2log2(an)=2-n
Tn=(3-n)n/2

cn=bn/an=2^(3-n)-4n/2^n>0

s(cn)=8-2^(3-n)-...

由于:4n/2^n-4/2^n>=0,所以s(cn)<=s(c1)=4

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