Example 2 Let $A B C D E F$ be a regular hexagon. A frog starts at vertex $A$ and can jump to one of the two adjacent vertices at each step. If it reaches point $D$ within 5 jumps, it stops jumping. If it does not reach point $D$ within 5 jumps, it stops after 5 jumps. How many different jumping sequences are possible from the start to the stop?

The number of different jumping sequences possible is $\qquad$.
As shown in Figure 1-1, it is clear that the frog cannot reach point $D$ by jumping 1 time, 2 times, or 4 times. Therefore, the frog's jumping methods are only in two scenarios.
(1) The frog reaches point $D$ by jumping 3 times, in which case there are only 2 ways to jump.
(2) The frog stops after jumping 5 times in total. In this case, the number of ways to jump 3 times (which definitely does not reach point $D$) is $2^{3}-2$ (each jump has 2 ways, and jumping 3 times has a total of $2^{3}$ ways, of which 2 ways reach point $D$ and should be excluded), and the number of ways for the last 2 jumps is $2^{2}$. Therefore, the number of ways the frog stops after jumping 5 times in total is $\left(2^{3}-2\right) \cdot 2^{2}=24$.

From (1) and (2), we know that the frog has a total of $2+24=26$ ways to jump.