Example 5 Let $A B C D E F$ be a hexagon, and a frog starts at vertex $A$. It 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?
(1997, National High School Mathematics Competition)
Analysis: Grasping the possible scenarios of the frog's jumping is the key to solving the problem.

Solution: According to the conditions, the frog's jumping methods can only result in two situations:
(1) Jump 3 times to reach point $D$, with 2 ways of jumping.
(2) Stop after 5 jumps (the first 3 jumps do not reach point $D$). Note that among the $2^{3}$ ways of the first 3 jumps, 2 ways reach point $D$, so there are $2^{3}-2=6$ ways for the first 3 jumps, and the last 2 jumps have $2^{2}$ ways, thus there are $6 \times 2^{2}=24$ ways.

From (1) and (2), we know that there are a total of $2+24=26$ ways of jumping.