Question

设S1=1+

1

12

+

1

22

，S2=1+

1

22

+

1

32

，S3=1+

1

32

+

1

42

，…，Sn=1+

1

n2

+

1

(n+1)2

．
设S=

S1

+

S2

+…+

Sn

，则S=

 

 （用含n的代数式表示，其中n为正整数）．

Answer 1

分析：由S_n=1+

1

n2

+

1

(n+1)2

=

n2(n+1)2+(n+1)2+n2

n2(n+1)2

=

[n(n+1)]2+2n2+2n+1

[n(n+1)]2

=

[n(n+1)+1]2

[n(n+1)]2

，求

Sn

，得出一般规律．

解答：解：∵S_n=1+

1

n2

+

1

(n+1)2

=

n2(n+1)2+(n+1)2+n2

n2(n+1)2

=

[n(n+1)]2+2n2+2n+1

[n(n+1)]2

=

[n(n+1)+1]2

[n(n+1)]2

，
∴

Sn

=

n(n+1)+1

n(n+1)

=1+

1

n(n+1)

=1+

1

n

-

1

n+1

，
∴S=1+1-

1

2

+1+

1

2

-

1

3

+…+1+

1

n

-

1

n+1

=n+1-

1

n+1

=

(n+1)2-1

n+1

=

n2+2n

n+1

．
故答案为：

n2+2n

n+1

．

点评：本题考查了二次根式的化简求值．关键是由S_n变形，得出一般规律，寻找抵消规律．