Question

設Sn=

1

2

+

1

6

+

1

12

+…+

1

n(n+1)

(n∈N*)，且Sn+1•Sn+2=

3

4

，則n的值是

5

．

Answer 1

分析：將通項裂項

1

n(n+1)

=

1

n

-

1

n+1

，求出Sn=

n

n+1

后，代入S _n+1•S_n+2 解出n即可．

解答：解：∵

1

n(n+1)

=

1

n

-

1

n+1

，∴Sn=(1-

1

2

)+ (

1

2

-

1

3

)+…+（

1

n

-

1

n+1

）=1-

1

n+1

=

n

n+1

∴S _n+1•S_n+2=

n+1

n+2

•

n+2

n+3

=

n+1

n+3

=

3

4

，解得n=5
故答案為：5．

點評：本題考查裂項法數列求和，常用于異分母分式形式．