3. Given the sequence $x_{n}$, satisfying $(n+1) x_{n+1}=x_{n}+n$, and $x_{1}=2$, then $x_{2009}=$ $\qquad$
3. $\frac{2009!+1}{2009!}$.

Solution: From $(n+1) x_{n+1}=x_{n}+n$, we derive $x_{n+1}-1=$ $\frac{x_{n}-1}{n+1}$, thus we have
$$
\begin{aligned}
& x_{n+1}-1=\frac{x_{n}-1}{n+1}=\frac{x_{n-1}-1}{(n+1) n}=\frac{x_{n-2}-1}{(n+1) n(n-1)} \\
=\cdots= & \frac{x_{1}-1}{(n+1) n(n-1) \cdots 2}=\frac{1}{(n+1)!} .
\end{aligned}
$$

Therefore, $x_{n+1}=\frac{1}{(n+1)!}+1$. Hence, we get $x_{2009}=$
$$
\frac{2009!+1}{2009!} \text {. }
$$