问题 解答题
已知数列{an}中a1=1,an+1=
an
2an+1
(n∈N+).
(1)求证:数列{
1
an
}
为等差数列;
(2)设bn=an•an+1(n∈N+),数列{bn}的前n项和为Sn,求满足Sn
1005
2012
的最小正整数n.
答案

(1)证明:由a1=1与an+1=

an
2an+1
得an≠0,
1
an+1
=
2an+1
an
=2+
1
an

所以对∀n∈N+

1
an+1
-
1
an
=2为常数,

{

1
an
}为等差数列;

(2)由(1)得

1
an
=
1
a1
+2(n-1)=2n-1,

bn=anan+1=

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

所以Sn=b1+b2+…+bn=

1
2
(1-
1
3
)+
1
2
(
1
3
-
1
5
)+…+
1
2
(
1
2n-1
-
1
2n+1
)=
1
2
(1-
1
2n+1
)
=
n
2n+1

Sn

1005
2012
n
2n+1
1005
2012
,得n>
1005
2
=502
1
2

所以满足Sn

1005
2012
的最小正整数n=503.

多项选择题
单项选择题