问题 解答题
已知函数f(x)=
x
x+3
,数列{an}满足a1=1,an+1=f(an)(n∈N+).
(1)求数列{an}的通项公式an
(2)若数列{bn}满足bn=
1
2
anan+13nSn=b1+b2+…+bn
,求Sn
答案

(1)由已知,an+1=

an
an+3
,所以
1
an+1
=
3
an
+1,

1
an+1
+
1
2
=3(
1
an
+
1
2
),

∴数列{

1
an
+
1
2
}是以1+
1
2
=
3
2
为首项,以3为公比的等比数列.

1
an
+
1
2
=
3
2
•3 n-1=
3n
2
1
an
=
3n-1
2

所以an=

2
3n-1

(2)bn=

1
2
anan+13n=
2•3n
(3n-1)(3n+1-1)
=
1
3n-1
-
1
3n+1-1

Sn=b1+b2+…+bn=

1
31-1
-
1
32-1
+(
1
32-1
-
1
33-1
)+…+(
1
3n-1
-
1
3n+1-1
)=
1
2
-
1
3n+1-1

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