问题 解答题
已知数列{an}为等差数列,{an}的前n项和为Sn,a1+a3=
3
2
,S5=5
(1)求数列{an}的通项公式;
(2)若数列{bn}满足anbn=
1
4
,Tn=b1b2+b2b3+b3b4+…+bnbn+1,求Tn
答案

(1)由a1+a3=

3
2
,S5=5,得
2a1+2d=
3
2
5a1+
5×4
2
d=5

解得a1=

1
2
,d=
1
4

an=

n+1
4

(2)∵an=

n+1
4
,∴bn=
1
n+1

bnbn+1=

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

∴Tn=b1b2+b2b3+b3b4+…+bnbn+1=(

1
2
-
1
3
)+(
1
3
-
1
4
)+…+(
1
n+1
-
1
n+2
)
=
1
2
-
1
n+2
=
n
2(n+2)

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