Question

先填空后计算：
①

1

n

- 

1

n+1

=

 

．

1

n+1

-

1

n+2

=

 

．

1

n+2

-

1

n+3

=

 

．
②计算：

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，再进行加减运算；
②根据题意得出规律：

1

n(n+1)

=

1

n

- 

1

n+1

；将每一项展开，再合并计算即可．

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)

．

点评：本题考查了分式的加减运算，先找出规律是解决此题的关键．