已知{an}是等差数列,其前n项和为Sn,已知a3=11,S9=153,(1)求数列{an}的通项公式;(2)设bn=2an,证明:{bn}是等比数列,并求其前n项和An.(3)设cn=1anan+1

题目简介

已知{an}是等差数列,其前n项和为Sn,已知a3=11,S9=153,(1)求数列{an}的通项公式;(2)设bn=2an,证明:{bn}是等比数列,并求其前n项和An.(3)设cn=1anan+1

题目详情

已知{an}是等差数列,其前n项和为Sn,已知a3=11,S9=153,
(1)求数列{an}的通项公式;
(2)设bn=2an,证明:{bn}是等比数列,并求其前n项和An
(3)设cn=
1
anan+1
,求其前n项和Bn
题型:解答题难度:中档来源:不详

答案

(1)∵{an}是等差数列,a3=11,S9=153,
∴9a5=153,
∴a5=17,
∴其公差d=
a5-a3
5-3
=3,
∴an=a5+(n-5)×d=17+(n-5)×3=3n+2;
(2)∵bn=2an,an=3n+2,
bn+1
bn
=2an+1-an=2d=23=8,且b1=25=32,
∴{bn}是以32为首项,8为公比的等比数列,
∴其前n项和An=class="stub"32
7
(8n-1);
(3)∵an=3n+2,
class="stub"1
anan+1
=class="stub"1
(3n+2)(3n+5)
=class="stub"1
3
class="stub"1
3n+2
-class="stub"1
3n+5
),
∴Bn=class="stub"1
3
[(class="stub"1
5
-class="stub"1
8
)+(class="stub"1
8
-class="stub"1
11
)+…+(class="stub"1
3n+2
-class="stub"1
3n+5
)]
=class="stub"1
3
class="stub"1
5
-class="stub"1
3n+5

=class="stub"n
15n+25

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