已知等差数列{an}的公差d>0,其前n项和为Sn,若S3=12,且2a1,a2,1+a3成等比数列.(I)求{an}的通项公式;(II)记bn=1anan+1(n∈N*),求数列{bn}的前n项和T

题目简介

已知等差数列{an}的公差d>0,其前n项和为Sn,若S3=12,且2a1,a2,1+a3成等比数列.(I)求{an}的通项公式;(II)记bn=1anan+1(n∈N*),求数列{bn}的前n项和T

题目详情

已知等差数列{an}的公差d>0,其前n项和为Sn,若S3=12,且2a1,a2,1+a3成等比数列.
(I)求{an}的通项公式;(II)记bn=
1
anan+1
(n∈N*)
,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

(I)由题得:
2a1(a3+1)=a22
a1+a2+a3=12

a1(a1+2d+1) =8
a1+d=4
,得d2+d-12=0.
∵d>0,∴d=3,a1=1.
∴{an}的通项公式an=1+3(n-1)=3n-2.
(II)∵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=b1+b2+b3+…+bn
=class="stub"1
3
[(1-class="stub"1
4
)+(class="stub"1
4
-class="stub"1
7
)+…+(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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