∵an=

1

n2+2n

=

1

n(n+2)

 =

1

2

(

1

n

-

1

n+2

 )，

∴S_n=a₁+a₂+a₃+…+a_n

=

1

2

(1-

1

3

) +

1

2

(

1

2

-

1

4

)+

1

2

( 

1

3

-

1

5

)+…+

1

2

(

1

n

-

1

n+2

)

=

1

2

(1+

1

2

-

1

n+1

-

1

n+2

)

=

3

4

-

2n+3

2(n+1)(n+2)

．

故答案为：

3

4

-

2n+3

2(n+1)(n+2)

．