Question

將(1-

1

x2

)n(n∈Nn)的展開(kāi)式中x^-4的系數(shù)記為a_n，則

1

a2

+

1

a3

+…+

1

a2010

=

 

．

Answer 1

分析：利用二項(xiàng)展開(kāi)式的通項(xiàng)公式求出展開(kāi)式中x^-4的系數(shù)a_n，求其倒數(shù)并將其裂成兩項(xiàng)的差，利用裂項(xiàng)法求出和．

解答：解：an=

C

2 n

=

n(n-1)

2

∴

1

an

=

2

n(n-1)

=2(

1

n-1

-

1

n

)
∴

1

a2

+

1

a3

+…+

1

a2010

=2[(1-

1

2

)+(

1

2

-

1

3

)+(

1

3

-

1

4

)+…+(

1

2009

-

1

2010

)]
=2(1-

1

2010

)
=

2009

1005

故答案為

2009

1005

點(diǎn)評(píng)：本題考查利用二項(xiàng)展開(kāi)式的通項(xiàng)公式解決二項(xiàng)展開(kāi)式的特定項(xiàng)問(wèn)題；通過(guò)裂項(xiàng)法求數(shù)列的前n項(xiàng)和．