Question

数列{a_n}中，a1=

1

2

，  an+1=

nan

(n+1)(nan+1)

(n∈N*)，则数列{a_n}的前2012项的和为

2012

2013

．

Answer 1

分析：由已知可得，

1

(n+1)an+1

=

nan+1

nan

=

1

nan

+1即

1

(n+1)an+1

-

1

nan

=1，

1

a1

=2，可得数列{

1

nan

}是以2为首项，以1为公差的等差数列，利用等差数列的通项公式可求

1

nan

，进而可求a_n，然后利用裂项求和即可求解

解答：解：∵a1=

1

2

，  an+1=

nan

(n+1)(nan+1)

(n∈N*)
∴

1

(n+1)an+1

=

nan+1

nan

=

1

nan

+1
∴

1

(n+1)an+1

-

1

nan

=1
∵a1=

1

2

∴

1

a1

=2
∴数列{

1

nan

}是以2为首项，以1为公差的等差数列
∴

1

nan

=2+(n-1)×1=n+1
∴an=

1

n(n+1)

=

1

n

-

1

n+1

∴S2012=1-

1

2

+

1

2

-

1

3

+…+

1

2012

-

1

2013

=1-

1

2013

=

2012

2013

故答案为：

2012

2013

点评：本题主要考查了利用数列的递推公式求解数列的 和，解题的关键是构造等差数列求出数列的通项公式，及裂项求和方法的应用．