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. From the given, we have
$$
a_{n+1}+\frac{1}{2}=\frac{n+2}{n}\left(a_{n}+\frac{1}{2}\right) \text {. }
$$

Let $b_{n}=a_{n}+\frac{1}{2}$. Then
$$
b_{n+1}=\frac{n+2}{n} b_{n} \text {. }
$$

Thus, $\frac{b_{n+1}}{b_{1}}=\frac{b_{n+1}}{b_{n}} \cdot \frac{b_{n}}{b_{n-1}} \cdots \cdots \frac{b_{2}}{b_{1}}$
$$
\begin{array}{l}
=\frac{n+2}{n} \cdot \frac{n+1}{n-1} \cdot \frac{n}{n-2} \cdots \cdots \frac{4}{2} \cdot \frac{3}{1} \\
=\frac{(n+2)(n+1)}{2} .
\end{array}
$$

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

[Note] Alternatively, the above can also be calculated as follows:

[The note is incomplete in the original text, so it is left as is.]