设数列{an}的前n项和为Sn,已知Sn=2an-2n+1(n∈N*).(1)求数列{an}的通项公式;(2)设数列{cn}满足cn=1log2(ann+1)+3(n∈N*),Tn=c1c2+c2c3

题目简介

设数列{an}的前n项和为Sn,已知Sn=2an-2n+1(n∈N*).(1)求数列{an}的通项公式;(2)设数列{cn}满足cn=1log2(ann+1)+3(n∈N*),Tn=c1c2+c2c3

题目详情

设数列{an}的前n项和为Sn,已知Sn=2an-2n+1(n∈N*).
(1)求数列{an}的通项公式;
(2)设数列{cn}满足cn=
1
log2(
an
n+1
)+3
(n∈N*)
,Tn=c1c2+c2c3+c3c4+…+cncn+1,若对一切n∈N*不等式4mTn>cn恒成立,求实数m的取值范围.
题型:解答题难度:中档来源:新余一模

答案

(1)当n=1时,a1=4(1分)
当n≥2时,an=Sn-Sn-1=2an-2an-1-2n⇒an=2an-1+2n(2分)
an
2n
=
an-1
2n-1
+1

{
an
2n
}
是首项为2,公差为1的等差数列(3分)
an
2n
=2+n-1⇒an=(n+1)•2n
(5分)
(2)cn=class="stub"1
n+3
cncn+1=class="stub"1
n+3
•class="stub"1
n+4
=class="stub"1
n+3
-class="stub"1
n+4
(7分)Tn=class="stub"1
4
-class="stub"1
5
+class="stub"1
5
-class="stub"1
6
+class="stub"1
6
-class="stub"1
7
++class="stub"1
n+3
-class="stub"1
n+4
=class="stub"1
4
-class="stub"1
n+4
=class="stub"n
4(n+4)
(9分)
4mTn>cn对一切n∈N*恒成立,则m>
(n+4)
n(n+3)
(11分)
class="stub"n+4
n(n+3)
=class="stub"n+4
n2+3n
=class="stub"n+4
(n+4)2-5(n+4)+4
=class="stub"1
(n+4)+class="stub"4
n+4
-5
≤class="stub"5
4
(13分)
m>class="stub"5
4
(14分)

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