设递增等比数列{an}的前n项和为Sn,且a2=3,S3=13,数列{bn}满足b1=a1,点P(bn,bn+1)在直线x-y+2=0上,n∈N*.(Ⅰ)求数列{an},{bn}的通项公式;(Ⅱ)设c

题目简介

设递增等比数列{an}的前n项和为Sn,且a2=3,S3=13,数列{bn}满足b1=a1,点P(bn,bn+1)在直线x-y+2=0上,n∈N*.(Ⅰ)求数列{an},{bn}的通项公式;(Ⅱ)设c

题目详情

设递增等比数列{an}的前n项和为Sn,且a2=3,S3=13,数列{bn}满足b1=a1,点P(bn,bn+1)在直线x-y+2=0上,n∈N*
(Ⅰ)求数列{an},{bn}的通项公式;
(Ⅱ)设cn=
bn
an
,数列{cn}的前n项和Tn,若Tn>2a-1恒成立(n∈N*),求实数a的取值范围.
题型:解答题难度:中档来源:不详

答案

(Ⅰ)∵递增等比数列{an}的前n项和为Sn,且a2=3,S3=13,
a2=3
S3=a1+a2+a3=13

解得q=3或q=class="stub"1
3

∵数列{an}为递增等比数列,所以q=3,a1=1.
∴{an}是首项为1,公比为3的等比数列.
an=3n-1.…(3分)
∵点P(bn,bn+1)在直线x-y+2=0上,
∴bn+1-bn=2.
∴数列{bn}是首项为1,公差为2的等差数列.
∴bn=1+(n-1)•2=2n-1.…(5分)
(Ⅱ)∵cn=
bn
an
=class="stub"2n-1
3n-1

Tn=class="stub"1
30
+class="stub"3
31
+class="stub"5
32
+…+class="stub"2n-1
3n-1

class="stub"1
3
Tn=class="stub"1
3
+class="stub"3
32
+class="stub"5
33
+…+class="stub"2n-3
3n-1
+class="stub"2n-1
3n
,…(7分)
两式相减得:
class="stub"2
3
Tn=class="stub"1
3
+class="stub"2
3
+class="stub"2
32
+…+class="stub"2
3n-1
-class="stub"2n-1
3n

=1+2×
class="stub"1
3
[1-(class="stub"1
3
)n-1]
1-class="stub"1
3
-class="stub"2n-1
3n

=2-(class="stub"1
3
)n-1-class="stub"2n-1
3n
.…(8分)
所以Tn=3-class="stub"1
2•3n-2
-class="stub"2n-1
2•3n-1
=3-class="stub"n+1
3n-1
.…(9分)
Tn+1-Tn=3-class="stub"n+2
3n
-3+class="stub"n+1
3n-1
=class="stub"2n+1
3n
>0
,…(10分)
∴Tn≥T1=1.
若Tn>2a-1恒成立,则1>2a-1,
解得a<1.
∴实数a的取值范围{a|a<1}.…(12分)

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