已知数列{an}的前n项和为Sn,a1=1,Sn+1=4an+1,设bn=an+1-2an.(Ⅰ)证明数列{bn}是等比数列;(Ⅱ)数列{cn}满足cn=1log2bn+3(n∈N+),设Tn=c1c

题目简介

已知数列{an}的前n项和为Sn,a1=1,Sn+1=4an+1,设bn=an+1-2an.(Ⅰ)证明数列{bn}是等比数列;(Ⅱ)数列{cn}满足cn=1log2bn+3(n∈N+),设Tn=c1c

题目详情

已知数列{an}的前n项和为Sn,a1=1,Sn+1=4an+1,设bn=an+1-2an
(Ⅰ)证明数列{bn}是等比数列;
(Ⅱ)数列{cn}满足cn=
1
log2bn+3
(n∈N+),设Tn=c1c2+c2c3+c3c4+…+cncn+1,若对一切n∈N+不等式4mTn>(n+2)cn恒成立,求实数m的取值范围.
题型:解答题难度:中档来源:东城区二模

答案

证明:(Ⅰ)由于Sn+1=4an+1,①
当n≥2时,Sn=4an-1+1.②
①-②得an+1=4an-4an-1.
所an+1-2an=2(an-2an-1).
又bn=an+1-2an,
所以bn=2bn-1.
因为a1=1,且a1+a2=4a1+1,
所以a2=3a1+1=4.
所以b1=a2-2a1=2.
故数列{bn}是首项为2,公比为2的等比数列.
(Ⅱ)由(Ⅰ)可知bn=2n,则cn=class="stub"1
log2bn+3
=class="stub"1
n+3

∴Tn=c1c2+c2c3+c3c4+…+cncn+1
=class="stub"1
4×5
+class="stub"1
5×6
+class="stub"1
6×7
+…+class="stub"1
(n+3)(n+4)

=class="stub"1
4
-class="stub"1
n+4

=class="stub"n
4(n+4)

由4mTn>(n+2),得class="stub"mn
n+4
class="stub"n+2
n+3

即m>
(n+4)(n+2)
n(n+3)

所以m>
n2+6n+8
n2+3n

所以m>1+class="stub"3n+8
n2+3n
=1+class="stub"3
n+3
+class="stub"8
n2+3n

设f(x)=1+class="stub"3
x+3
+class="stub"8
x2+3x
,x≥1.
可知f(x)在[1,+∞)为减函数,又f(1)=class="stub"15
4

则当n∈N时,有f(n)≤f(1).
所以∴m>class="stub"15
4

故当m>class="stub"15
4
.时,4mTn>(n+2)cn恒成立.

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