∵1+2+…+n=

1

2

n(n+1)

∴an=

1

n+1

+

2

n+1

+…+

n

n+1

=

n

2

∴bn=

2

n 2 • n+1 2

=8(

1

n

-

1

n+1

)

∴数列{b_n}的前n项和Sn=8[(1-

1

2

)+(

1

2

-

1

3

)+(

1

3

-

1

4

)+…+(

1

n

-

1

n+1

)]=8(1-

1

n+1

)=

8n

n+1