Example 4.14 Let $A_{n+1}$ be a convex $(n+1)$-gon in the plane $(n \geqslant 2)$. Use $n-2$ non-intersecting diagonals to dissect $A_{n+1}$ into $n-1$ triangles (called a triangular dissection of $A_{n+1}$). How many different dissection methods are there?
Let $h_{n}$ denote the number of different triangular subdivisions of $A_{n+1}$. It is easy to see that $h_{2}=1$, $h_{3}=2$. Suppose $n \geqslant 4$, and select an edge in $A_{n+1}$, denoted as $l$. Then the $h_{n}$ different triangular subdivisions of $A_{n+1}$ can be divided into the following two categories:
(1) The triangular subdivision $T$ containing $l$ has only one diagonal.
Since there are two such $T$s, and removing $T$ results in a convex $n$-gon, which has $h_{n-1}$ triangular subdivisions. Therefore, by the principle of addition, the number of triangular subdivisions in this category is $2 \cdot h_{n-1}$.
(2) The triangular subdivision $T$ containing $l$ has two diagonals.

Since removing $T$ results in two convex polygons $B_{1}$ and $B_{2}$ (as shown in Figure 4.1), and the sum of the number of sides of $B_{1}$ and $B_{2}$ is $n+1+2-1=n+2$, if $B_{1}$ has $k+1(2 \leqslant k \leqslant n-2)$ sides, then $B_{2}$ has $n+1-k$ sides, and the triangular subdivisions of $B_{1}$ and $B_{2}$ are $h_{k}$ and $h_{n-k}$ respectively. Therefore, the number of triangular subdivisions in this category is $\sum_{k=2}^{n-2} h_{k} h_{n-k}$.
By the principle of addition, we have
$$
h_{n}=2 h_{n-1}+\sum_{k=2}^{n-2} h_{k} h_{n-k} \quad(n \geqslant 4) .
$$

Let $h_{1}=1$, then
$$
h_{n}=h_{1} \cdot h_{n-1}+\sum_{k=2}^{n-2} h_{k} h_{n-k}+h_{n-1} \cdot h_{1},
$$

Thus,
$$
h_{n}=\sum_{k=1}^{n-1} h_{k} h_{n-k} \quad(n \geqslant 4) \text {. }
$$

It is easy to see that the above formula still holds when $n=2,3$. Let $y=\sum_{n=1}^{\infty} h_{n} t^{n}$, then
$$
\begin{aligned}
y^{2} & =\left(\sum_{k=1}^{\infty} h_{k} t^{k}\right)\left(\sum_{j=1}^{\infty} h_{j} t^{j}\right) \\
& =\sum_{n=2}^{\infty}\left(\sum_{k=1}^{n-1} h_{k} h_{n-k}\right) t^{n} \\
& =\sum_{n=2}^{\infty} h_{n} t^{n}=y-t,
\end{aligned}
$$

Therefore,
$$
\begin{array}{c}
y^{2}-y+t=0, \\
y=\frac{1}{2}(1+\sqrt{1-4 t}) \text { or } y=\frac{1}{2}(1-\sqrt{1-4} t) .
\end{array}
$$

Since $h_{n}>0$, and
$$
\begin{aligned}
\sqrt{1-4 t} & =1-\sum_{n=1}^{\infty} \frac{1}{2^{2 n-1}} \cdot \frac{1}{n}\binom{2 n-2}{n-1}(4 t)^{n} \\
& =1-2 \sum_{n=1}^{\infty} \frac{1}{n}\binom{2 n-2}{n-1} t^{n}, \\
\sum_{n=1}^{\infty} h_{n} t^{n} & =y=\frac{1}{2}(1-\sqrt{1-4 t}) \\
& =\sum_{n=1}^{\infty} \frac{1}{n}\binom{2 n-2}{n-1} t^{n},
\end{aligned}
$$

Therefore, $h_{n}=\frac{1}{n}\binom{2 n-2}{n-1}$, which means $h_{n}$ is the Catalan number $C_{n}$.