∵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

．