（I）由条件得，an+1=

an

3an+1

．

∴

1

an+1

=

1

an

+3⇒

1

an+1

-

1

an

=3．

∴数列{

1

an

}是首项为

1

a1

=1，公差d=3的等差数列．

∴

1

an

=1+（n-1）×3=3n-2．

故a_n=

1

3n-2

．

（II）∵a_na_n+1=

1

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

=

1

3

（

1

3n-2

-

1

3n+1

）．

∴S_n═a₁a₂+a₂a₃+..a_na_n+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

．