∵观察下列等式：

1+ 1 12 + 1 22

=1+

1

1

-

1

1+1

=1

1

2

1+ 1 22 + 1 32

=1+

1

2

+

1

2+1

=1

1

6

1+ 1 32 + 1 42

=1+

1

3

-

1

3+1

=1

1

12

…

∴第n个等式是

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

=1+

1

n

-

1

n+1

=1+

1

n(n+1)

，

故答案为：

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

=1+

1

n

-

1

n+1

=1+

1

n(n+1)

．