Example 5 For a quadrilateral, there can be 2 ways to dissect it into triangles, $a_{4}=2$ (Figure 15); for a pentagon, there can be 5 ways to dissect it into triangles, $a_{5}=$ 5 (Figure 16). How many ways are there to dissect a hexagon into triangles?
Solution: Consider the triangles containing the edge $A_{1} A_{2}$, using $\triangle A_{1} A_{2} A_{k}(k=3,4,5,6)$ as a standard, the figure can be divided into 4 categories.
(1) When the triangulation graph must contain $\triangle A_{1} A_{2} A_{3}$, we can take out $\triangle A_{1} A_{2} A_{3}$, leaving the pentagon $A_{3} A_{4} A_{5} A_{6} A_{1}$, which has 5 triangulation methods (Figure 17).
(2) When the triangulation graph must contain $\triangle A_{1} A_{2} A_{4}$, we can take out $\triangle A_{1} A_{2} A_{4}$, leaving the triangle $\triangle A_{2} A_{3} A_{4}$ and the quadrilateral $A_{1} A_{4} A_{5} A_{6}$, which have 2 triangulation methods (Figure 18).
(3) When the triangulation graph must contain $\triangle A_{1} A_{2} A_{5}$, the situation is the same as (2), with 2 triangulation methods (Figure 19).
(4) When the triangulation graph must contain $\triangle A_{1} A_{2} A_{6}$, the situation is the same as (1), with 5 triangulation methods (Figure 20).
In total, there are $5+2+2+5=14$ triangulation methods.
Analysis: If we denote the number of triangulation methods for an $n$-sided polygon as $a_{n}$, and set $a_{2}=a_{3}=1$, then by the same method of taking out 1 $\triangle A_{1} A_{2} A_{k}(3 \leqslant k \leqslant n)$ (Figure 21), we get 1 $(k-1)$-sided polygon and 1 $(n-k+2)$-sided polygon, with $a_{k-1} a_{n-k+2}$ triangulation methods. Taking all $k=3,4, \cdots, n$, we get (the recursive formula)
$$
a_{n}=a_{2} a_{n-1}+a_{3} a_{n-2}+\cdots+a_{n-1} a_{2} .
$$