{an}为等差数列,公差d>0,Sn是数列{an}前n项和,已知a1a4=27,S4=24.(1)求数列{an}的通项公式an;(2)令bn=1anan+1,求数列{bn}的前n项和Tn.-数学

题目简介

{an}为等差数列,公差d>0,Sn是数列{an}前n项和,已知a1a4=27,S4=24.(1)求数列{an}的通项公式an;(2)令bn=1anan+1,求数列{bn}的前n项和Tn.-数学

题目详情

{an}为等差数列,公差d>0,Sn是数列{an}前n项和,已知a1a4=27,S4=24.
(1)求数列{an}的通项公式an
(2)令bn=
1
anan+1
,求数列{bn}的前n项和Tn
题型:解答题难度:中档来源:不详

答案

(1)S4=
4(a1+a4)
2
=24
,∴a1+a4=12
又a1a4=27,d>0,∴a1=3,a4=9,
∴9=3+3d,解得d=2,
∴an=2n+1.
(2)bn=class="stub"1
anan+1
=class="stub"1
(2n+1)(2n+3)
=class="stub"1
2
(class="stub"1
2n+1
-class="stub"1
2n+3
)

Tn=class="stub"1
2
[(class="stub"1
3
-class="stub"1
5
)+(class="stub"1
5
-class="stub"1
7
)+…+(class="stub"1
2n+1
-class="stub"1
2n+3
)]=class="stub"1
2
(class="stub"1
3
-class="stub"1
2n+3
)

=class="stub"n
6n+9

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