问题 解答题
已知公差d不为0的等差数列{an}中,a1=1,且a1,a3,a7成等比数列.
(1)求通项an及前n项和Sn
(2)若有一新数列{bn},且bn=
1
anan+1
,求数列{bn}的前n项和Tn
答案

(1)∵a1,a3,a7成等比数列,

a32=a1a7

即(1+2d)2=a1+6d,

4d2-2d=0,d=

1
2
,或d=0(舍去).

∴数列的通项公式an=1+

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

前n项和SnSn=

n(1+
1
2
n+
1
2
)
2
=
1
4
n2+
3
4
n.

(2)由(1)得an=

n+1
2

an+1=

n+2
2

bn=

1
anan+1
=
4
(n+1)(n+2)
=4(
1
n+1
-
1
n+2
),

Tn=4(

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

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