问题 解答题
(文)数列{an}满足an+1=
n+2
n
an
(n∈N*),且a1=1.(1)求通项an;(2)记bn=
1
an
,数列{bn}的前n项和为Sn,求Sn
答案

解(1)∵an+1=

n+2
n
an

an+1
an
=
n+2
n

∵a1=1

a2
a1
=
3
1
a3
a2
=
4
2
an
an-1
=
n+1
n-1

以上n-1个式子相乘可得,

a2
a1
a3
a2
an
an-1
=
3
1
×
4
2
×
5
3
n-1
n-3
×
n
n-2
×
n+1
n-1

an
a1
=
n(n+1)
1×2

∴an=

n(n+1)
2

(2)∵bn=

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

Sn=2(1-

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

单项选择题 A1/A2型题
单项选择题