Question

先填空后計算：
①

1

n

- 

1

n+1

=

 

．

1

n+1

-

1

n+2

=

 

．

1

n+2

-

1

n+3

=

 

．
②計算：

1

n(n+1)

+

1

(n+1)(n+2)

+

1

(n+2)(n+3)

+…+

1

(n+2007)(n+2008)

．

Answer 1

分析：①先通分成同分母的分式，再進行加減運算；
②根據(jù)題意得出規(guī)律：

1

n(n+1)

=

1

n

- 

1

n+1

；將每一項展開，再合并計算即可．

解答：解：①原式=

n+1-n

n(n+1)

=

1

n(n+1)

；
原式=

n+2-n-1

(n+1)(n+2)

=

1

(n+1)(n+2)

；
原式=

n+3-n-2

(n+2)(n+3)

=

1

(n+2)(n+3)

；
②原式=

1

n

- 

1

n+1

+

1

n+1

-

1

n+2

+

1

n+2

-

1

n+3

+…+

1

n+2007

-

1

n+2008

=

1

n

-

1

n+2008

=

n+2008-n

n(n+2008)

=

2008

n(n+2008)

．

點評：本題考查了分式的加減運算，先找出規(guī)律是解決此題的關(guān)鍵．