(14) Given the sequence $\left\{a_{n}\right\}$ satisfies, $a_{1}=0, a_{n+1}=\frac{n+2}{n} a_{n}+\frac{1}{n}$, find the general term formula of the sequence $\left\{a_{n}\right\}$.
14 Solution 1 From the given, we have
$$
a_{n+1}+\frac{1}{2}=\frac{n+2}{n}\left(a_{n}+\frac{1}{2}\right) .
$$

Let $b_{n}=a_{n}+\frac{1}{2}$, then $b_{n+1}=\frac{n+2}{n} b_{n}$, hence
$$
\begin{aligned}
\frac{b_{n+1}}{b_{1}} & =\frac{b_{n+1}}{b_{n}} \cdot \frac{b_{n}}{b_{n-1}} \cdots \cdot \frac{b_{2}}{b_{1}} \\
& =\frac{n+2}{n} \cdot \frac{n+1}{n-1} \cdot \frac{n}{n-2} \cdots \cdot \frac{4}{2} \cdot \frac{3}{1} \\
& =\frac{(n+2)(n+1)}{2} .
\end{aligned}
$$

Since $b_{n}=a_{n}+\frac{1}{2}, b_{1}=\frac{1}{2}$, we have
$$
a_{n}=\frac{(n+1) n}{4}-\frac{1}{2}=\frac{(n-1)(n+2)}{4} \text {. }
$$

Solution 2 We can first calculate $a_{2}=1, a_{3}=\frac{5}{2}, a_{4}=\frac{9}{2}, \cdots$, and guess that $a_{n}=\frac{(n-1)(n+2)}{4}$, then prove it using mathematical induction.