Question

计算：

1

1×2×3×4

+

1

2×3×4×5

+

1

3×4×5×6

+…+

1

n(n+1)(n+2)(n+3)

．

Answer 1

考点：分式的加减法

专题：计算题

分析：根据规律

1

n(n+1)(n+2)(n+3)

=

1

3

[

1

n(n+1)(n+2)

-

1

(n+1)(n+2)(n+3)

]，把原式进行变形，计算即可得到结果．

解答：解：原式=

1

3

[

1

1×2×3

-

1

2×3×4

+

1

2×3×4

-

1

3×4×5

+…+

1

n(n+1)(n+2)

-

1

(n+1)(n+2)(n+3)

]=

1

3

（

1

1×2×3

-

1

(n+1)(n+2)(n+3)

）=

1

18

-

1

3(n+1)(n+2)(n+3)

．

点评：此题考查了分式的加减法，弄清拆项规律是解本题的关键．