1. If $x_{n+1}=\frac{n+2}{n} x_{n}+\frac{1}{n}, x_{1}>0$, find $x_{n}$.
$$
\begin{array}{l}
\quad x_{n+1}+\frac{1}{2}=\frac{n+2}{n}\left(x_{n}+\frac{1}{2}\right) \Rightarrow x_{n}=\frac{(n+1)!}{(n-1)!} \times \\
\left.\frac{1}{4}-\frac{1}{2}(n \geqslant 2) .\right)
\end{array}
$$

(Tip: I derived the fixed point $\alpha=-\frac{1}{2}$. We have
$$
\begin{array}{l}
\quad x_{n+1}+\frac{1}{2}=\frac{n+2}{n}\left(x_{n}+\frac{1}{2}\right) \Rightarrow x_{n}=\frac{(n+1)!}{(n-1)!} \times \\
\left.\frac{1}{4}-\frac{1}{2}(n \geqslant 2) .\right)
\end{array}
$$
)