Question

計(jì)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

考點(diǎn)：分式的加減法

專題：計(jì)算題

分析：根據(jù)規(guī)律

1

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

=

1

3

[

1

n(n+1)(n+2)

-

1

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

]，把原式進(jìn)行變形，計(jì)算即可得到結(jié)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)

．

點(diǎn)評(píng)：此題考查了分式的加減法，弄清拆項(xiàng)規(guī)律是解本題的關(guān)鍵．