数列{an}是首项a1=4的等比数列,且S3,S2,S4成等差数列,(1)求数列{an}的通项公式;(2)若bn=log2|an|,设Tn为数列{1bnbn+1}的前n项和,若Tn≤λbn+1对一切n

题目简介

数列{an}是首项a1=4的等比数列,且S3,S2,S4成等差数列,(1)求数列{an}的通项公式;(2)若bn=log2|an|,设Tn为数列{1bnbn+1}的前n项和,若Tn≤λbn+1对一切n

题目详情

数列{an}是首项a1=4的等比数列,且S3,S2,S4成等差数列,
(1)求数列{an}的通项公式;
(2)若bn=log2|an|,设Tn为数列{
1
bnbn+1
}
的前n项和,若Tn≤λbn+1对一切n∈N*恒成立,求实数λ的最小值.
题型:解答题难度:中档来源:不详

答案

(1)∵S3,S2,S4成等差数列
∴2S2=S3+S4即2(a1+a2)=2(a1+a2+a3)+a4
所以a4=-2a3
∴q=-2
an=a1qn-1=(-2)n+1
(2)bn=log2|an|=log22n+1=n+1
class="stub"1
bnbn+1
=class="stub"1
(n+1)(n+2)
=class="stub"1
n+1
-class="stub"1
n+2

Tn=(class="stub"1
2
-class="stub"1
3
)+(class="stub"1
3
-class="stub"1
4
)+…+(class="stub"1
n+1
-class="stub"1
n+2
)=class="stub"1
2
-class="stub"1
n+2

λ≥
Tn
bn+1
=class="stub"n
2(n+2)2
=class="stub"1
2
×class="stub"1
n+class="stub"4
n
+4

因为n+class="stub"4
n
≥4,所以class="stub"1
2
×class="stub"1
n+class="stub"4
n
+4
class="stub"1
16

所以λ最小值为class="stub"1
16

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