Problem 81

Solution:
- (I) Since $a_1=1$ and $a_{n+1}= \frac {a_n}{3a_n+1}$,
$\therefore a_2= \frac {a_1}{3a_1+1}= \frac {1}{4}$;
$a_3= \frac {a_2}{3a_2+1}= \frac { \frac {1}{4}}{ \frac {3}{4}+1}= \frac {1}{7}$, $a_4= \frac { \frac {1}{7}}{ \frac {3}{7}+1}= \frac {1}{10}$;
- (II) From (I), we can conjecture: $a_n= \frac {1}{3n-2}$.
Proof:
1) When $n=1$, $a_1=1$, the equation holds;
2) Assume when $n=k$, $a_k= \frac {1}{3k-2}$,
then when $n=k+1$, $a_{k+1}= \frac {a_k}{3a_k+1}= \frac { \frac {1}{3k-2}}{3× \frac {1}{3k-2}+1}= \frac {1}{3k+1}= \frac {1}{3(k+1)-2}$,
which means the equation also holds for $n=k+1$.
In conclusion, for any natural number $n\in\mathbb{N}^*$, $a_n= \frac {1}{3n-2}$.
Thus, the final answer is $\boxed{a_n= \frac {1}{3n-2}}$.