Question

數(shù)列{a_n}中，a1=

1

2

，  an+1=

nan

(n+1)(nan+1)

(n∈N*)，則數(shù)列{a_n}的前2012項(xiàng)的和為

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，可得數(shù)列{

1

nan

}是以2為首項(xiàng)，以1為公差的等差數(shù)列，利用等差數(shù)列的通項(xiàng)公式可求

1

nan

，進(jìn)而可求a_n，然后利用裂項(xiàng)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
∴數(shù)列{

1

nan

}是以2為首項(xiàng)，以1為公差的等差數(shù)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

故答案為：

2012

2013

點(diǎn)評：本題主要考查了利用數(shù)列的遞推公式求解數(shù)列的 和，解題的關(guān)鍵是構(gòu)造等差數(shù)列求出數(shù)列的通項(xiàng)公式，及裂項(xiàng)求和方法的應(yīng)用．