已知递增的等比数列{an}满足:a2+a3+a4=28,a3+2是a2与a4的等差中项.(1)求数列{an}的通项公式;(2)假设bn=an(an+1)(an+1+1),其数列{bn}的前n项和Tn,

题目简介

已知递增的等比数列{an}满足:a2+a3+a4=28,a3+2是a2与a4的等差中项.(1)求数列{an}的通项公式;(2)假设bn=an(an+1)(an+1+1),其数列{bn}的前n项和Tn,

题目详情

已知递增的等比数列{an}满足:a2+a3+a4=28,a3+2是a2与a4的等差中项.
(1)求数列{an}的通项公式;
(2)假设bn=
an
(an+1)(an+1+1)
,其数列{bn}的前n项和Tn,并解不等式Tn
127
390
题型:解答题难度:中档来源:不详

答案

(1)∵递增的等比数列{an}满足:a2+a3+a4=28,a3+2是a2与a4的等差中项,
∴2(a3+2)=a2+a4,a3=8,a2+a4=80,
a1q2=8
a1q+a1q3=20

解得a1=2,q=2,或a1=32,q=class="stub"1
2
(舍),
an=2n
(2)bn=
an
(an+1)(an+1+1)

=
2n
(2n+1)(2n+1+1)

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

∴Tn=class="stub"1
2+1
-class="stub"1
22+1
+class="stub"1
22+1
-class="stub"1
23-1
+…+class="stub"1
2n-1+1
-class="stub"1
2n+1
+class="stub"1
2n+1
-class="stub"1
2n+1+1

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

=class="stub"1
3
-class="stub"1
2n+1+1

∵Tn<class="stub"127
390

class="stub"1
3
-class="stub"1
2n+1+1
<class="stub"127
130
,∴2n+1<129,解得n≤6,
∴不等式Tn<class="stub"127
390
的解集为{1,2,3,4,5,6}.

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