已知等差数列{an}的前n项和为sn,且s3=12,2a1,a2,a3+1成公比大于1的等比数列.(1)求{an}的通项公式;(2)bn=1anan+1,求{bn}的前n项和Tn.-数学

题目简介

已知等差数列{an}的前n项和为sn,且s3=12,2a1,a2,a3+1成公比大于1的等比数列.(1)求{an}的通项公式;(2)bn=1anan+1,求{bn}的前n项和Tn.-数学

题目详情

已知等差数列{an}的前n项和为sn,且s3=12,2a1,a2,a3+1成公比大于1的等比数列.
(1)求{an}的通项公式;
(2)bn=
1
anan+1
,求{bn}的前n项和Tn
题型:解答题难度:中档来源:不详

答案

(1)∵2a1,a2,a3+1成公比大于1的等比数列
a22=2a1(a3+1)
(a1+d)2=2a1(a1+2d+1)
∵3a1+3d=12
联立①②可得,
a1=1
d=3
a1=8
d=-4

a2
2a1
1
a1=1
d=3
,an=1+3(n-1)=3n-2
(2)∵bn=class="stub"1
anan+1
=class="stub"1
(3n-2)(3n+1)
=class="stub"1
3
(class="stub"1
3n-2
-class="stub"1
3n+1
)

Tn=class="stub"1
3
(1-class="stub"1
4
+class="stub"1
4
-class="stub"1
7
+class="stub"1
7
-class="stub"1
10
+…+class="stub"1
3n-2
-class="stub"1
3n+1

=class="stub"1
3
(1-class="stub"1
3n+1
)
=class="stub"n
3n+1

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