试题

题目：

先填空后计算：
①

1

n

-

1

n+1

=

1

n(n+1)

1

n(n+1)

．

1

n+1

-

1

n+2

=

1

(n+1)(n+2)

1

(n+1)(n+2)

．

1

n+2

-

1

n+3

=

1

(n+2)(n+3)

1

(n+2)(n+3)

．
②计算：

1

n(n+1)

+

1

(n+1)(n+2)

+

1

(n+2)(n+3)

+…+

1

(n+2007)(n+2008)

．

答案

1

n(n+1)

1

(n+1)(n+2)

1

(n+2)(n+3)

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)

．

考点梳理

分式的加减法．

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