已知数列{an}的前n项和为Sn,且Sn=12an•an+1(n∈N*),其中a1=1,an≠0.(Ⅰ)求数列{an}的通项公式;(Ⅱ)设数列{bn}满足(2an-1)(2bn-1)=1,Tn为{bn

题目简介

已知数列{an}的前n项和为Sn,且Sn=12an•an+1(n∈N*),其中a1=1,an≠0.(Ⅰ)求数列{an}的通项公式;(Ⅱ)设数列{bn}满足(2an-1)(2bn-1)=1,Tn为{bn

题目详情

已知数列{an}的前n项和为Sn,且Sn=
1
2
anan+1(n∈N*)
,其中a1=1,an≠0.
(Ⅰ)求数列{an}的通项公式;
(Ⅱ)设数列{bn}满足(2an-1)(2bn-1)=1,Tn为{bn}的前n项和,求证:2Tn>log2(2an+1)n∈N.
题型:解答题难度:中档来源:朝阳区一模

答案

(Ⅰ)已知式即Sn=class="stub"1
2
anan+1
,故an+1=Sn+1-Sn=class="stub"1
2
an+1an+2-class="stub"1
2
anan+1

由条件知an+1≠0,所以an+2-an=2(n∈N*).
由于a1=S1=class="stub"1
2
a1a2
,且a1=1,故a2=2.
于是a2m-1=1+2(m-1)=2m-1,a2m=2+2(m-1)=2m,
所以an=n(n∈N*).
(Ⅱ)由(2an-1)(2bn-1)=1,得(2n-1)(2bn-1)=12bn=class="stub"2n
2n-1

bn=log2class="stub"2n
2n-1

从而Tn=b1+b2++bn=log2(class="stub"2
1
•class="stub"4
3
•class="stub"6
5
••class="stub"2n
2n-1
)

2Tn=2log2(class="stub"2
1
•class="stub"4
3
•class="stub"6
5
••class="stub"2n
2n-1
)
=log2(class="stub"2
1
•class="stub"4
3
•class="stub"6
5
••class="stub"2n
2n-1
)2

因此2Tn-log2(2an+1)=log2(class="stub"2
1
•class="stub"4
3
•class="stub"6
5
••class="stub"2n
2n-1
)2
-log2(2n+1)
=log2(class="stub"2
1
•class="stub"4
3
•class="stub"6
5
••class="stub"2n
2n-1
)2+log2class="stub"1
2n+1

=log2[(class="stub"2
1
•class="stub"4
3
•class="stub"6
5
••class="stub"2n
2n-1
)2•class="stub"1
2n+1
]

f(n)=(class="stub"2
1
•class="stub"4
3
•class="stub"6
5
••class="stub"2n
2n-1
)2•class="stub"1
2n+1

f(n+1)=(class="stub"2
1
•class="stub"4
3
•class="stub"6
5
••class="stub"2n
2n-1
•class="stub"2n+2
2n+1
)2•class="stub"1
2n+3

f(n+1)
f(n)
=class="stub"2n+1
2n+3
•(class="stub"2n+2
2n+1
)2=
(2n+2)2
(2n+3)(2n+1)
=
4n2+8n+4
4n2+8n+3
>1

注意到f(n)>0,所以f(n+1)>f(n).
特别地f(n)≥f(1)=class="stub"4
3
>1

从而2Tn-log2(2an+1)=log2f(n)>0.
所以2Tn>log2(2an+1).

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