Question

（理）S=

1

1×2×3

+

1

2×3×4

+…+

1

n(n+1)(n+2)

+…，则S=

1

4

-

1

2(n+1)(n+2)

．

Answer 1

分析：

1

n(n+1)(n+2)

=

1

2

[

1

n(n+1)

-

1

(n+1)(n+2)

]，利用裂项相消法可求和S．

解答：解：∵

1

n(n+1)(n+2)

=

1

2

[

1

n(n+1)

-

1

(n+1)(n+2)

]，
∴S=

1

2

[

1

1×2

-

1

2×3

+

1

2×3

-

1

3×4

+…+

1

n(n+1)

-

1

(n+1)(n+2)

]
=

1

2

[

1

2

-

1

(n+1)(n+2)

]=

1

4

-

1

2(n+1)(n+2)

，
故答案为：

1

4

-

1

2(n+1)(n+2)

．

点评：本题考查数列求和，裂项相消法对数列求和是高考考查重点，应熟练掌握．