Question

Sn=

1

1×2

+

1

2×3

+

1

3×4

+

1

4×5

+…+

1

n(n+1)

=

n

n+1

．

Answer 1

分析：由数列通项

1

n(n+1)

=

1

n

-

1

n+1

，想到利用裂项相消法求和．

解答：解：∵

1

n(n+1)

=

1

n

-

1

n+1

，
∴Sn=

1

1×2

+

1

2×3

+

1

3×4

+

1

4×5

+…+

1

n(n+1)

=(1-

1

2

)+(

1

2

-

1

3

)+(

1

3

-

1

4

)+…+(

1

n

-

1

n+1

)
=1-

1

n+1

=

n

n+1

．
故答案为：

n

n+1

．

点评：本题考查了数列的求和，训练了裂项相消法，属中低档题．