设数列{an}的前n项和为Sn,已知Sn=2an-2n+1(n∈N*).(1)设bn=an2n,求证:数列{bn}是等差数列:(2)设数列{cn}满足cn=1log2(ann+1)+1(n∈N*),T

题目简介

设数列{an}的前n项和为Sn,已知Sn=2an-2n+1(n∈N*).(1)设bn=an2n,求证:数列{bn}是等差数列:(2)设数列{cn}满足cn=1log2(ann+1)+1(n∈N*),T

题目详情

设数列{an}的前n项和为Sn,已知Sn=2an-2n+1(n∈N*).
(1)设bn=
an
2n
,求证:数列{bn}是等差数列:
(2)设数列{cn}满足cn=
1
log2(
an
n+1
) +1
(n∈N*),Tn=c1c2+c2c3+c3c4+…cncn+1,若对一切n∈N*不等式2mTn>Cn恒成立,实数m的取值范围.
题型:解答题难度:中档来源:不详

答案

(1)当n=1时:S1=a1=2a1-21|1,解得a1=4
当n≥2时
由Sn=2an-2n+1 …①
且Sn-1=2an-1-2n …②
①-②得:an=2an-2an-1-2n
有:an=2an-1+2n
an
2n
-
an-1
2n-1
=1

∴bn-bn-1=1,
b1=
a1
2
=2

故数列{bn}是以2为首项,以1为公差的等差数列.
(2)由(1)得:bn=1+2(n-1)=2n-1,
即an=(n+1)•2n.
Cn=class="stub"1
n+1

CnCn+1=class="stub"1
n+1
•class="stub"1
n+2
=class="stub"1
n+1
-class="stub"1
n+2

Tn=class="stub"1
2
-class="stub"1
n+2

由2mTn>cn,得:2m(class="stub"1
2
-class="stub"1
n+2
)>class="stub"1
n+1

m>class="stub"n+2
n(n+1)

又令f(n)=class="stub"n+2
n(n+1)

f(n+1)-f(n)=class="stub"n+3
(n+1)(n+2)
-class="stub"n+2
n(n+1)

=class="stub"1
n+1
(class="stub"n+3
n+2
-class="stub"n+2
n
)<0

故f(n)在n∈N*时单调递减,
f(n)<f(1)=class="stub"3
2

得m>class="stub"3
2

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