（1）由已知得，an+1=

an

3an+1

，整理得

1

an+1

-

1

an

=3

∴数列{

1

an

}是首项，公差的等差数列．

∴

1

an

=1+(n-1)×3=3n-2，

故an=

1

3n-2

(n∈N*)（6分）

（2）∵anan+1=

1

(3n-2)(3n+1)

=

1

3

(

1

3n-2

-

1

3n+1

)

S_n=a₁a₂+a₂a₃+…+a_na_n+1=

1

1×4

+

1

4×7

+…+

1

(3n-2)(3n+1)

=

1

3

[(1-

1

4

)+(

1

4

-

1

7

)+…+(

1

3n-2

-

1

3n+1

)]

=

1

3

(1-

1

3n+1

)=

n

3n+1

．（13分）