已知等比数列{an}的各项均为正数,且a1+2a2=1,a23=4a2a6.(1)求数列{an}的通项公式;(2)设bn=log2a1+log2a2+…+log2an,求数列{1bn}的前n项和.-高

题目简介

已知等比数列{an}的各项均为正数,且a1+2a2=1,a23=4a2a6.(1)求数列{an}的通项公式;(2)设bn=log2a1+log2a2+…+log2an,求数列{1bn}的前n项和.-高

题目详情

已知等比数列{an}的各项均为正数,且a1+2a2=1,a
23
=4a2a6
(1)求数列{an}的通项公式;
(2)设bn=log2a1+log2a2+…+log2an,求数列{
1
bn
}的前n项和.
题型:解答题难度:中档来源:不详

答案

(1)设等比数列{an}的公比为q,由a
23
=4a2a6得a
23
=4
a24

∴q2=class="stub"1
4
,由已知an>0,∴q=class="stub"1
2

由a1+2a2=1,得2a1=1,∴a1=class="stub"1
2

∴数列{an}的通项公式为an=class="stub"1
2n

(2)bn=log2a1+log2a2+…+log2an=-(1+2+…+n)=-
n(n+1)
2

class="stub"1
bn
=-class="stub"2
n(n+1)
=-2(class="stub"1
n
-class="stub"1
n+1
),
∴数列{class="stub"1
bn
}的前n项和=-2[(1-class="stub"1
2
)+(class="stub"1
2
-class="stub"1
3
)+…+(class="stub"1
n
-class="stub"1
n+1
)]=-class="stub"2n
n+1

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