5. Let $A B C D E F$ be a regular hexagon. A frog starts at vertex $A$, and each time it can randomly jump to one of the two adjacent vertices. 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 can the frog have from the start to the stop? $\qquad$
26
5. 【Analysis and Solution】As shown in the figure, 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 of the following two types:

The frog jumps 3 times to reach point $D$, with 2 ways to do so; the frog stops after a total of 5 jumps. In this case, the first 3 jumps (which definitely do not reach point $D$) have $\left(2^{3}-2\right)$ ways, and the last two jumps have $2^{2}$ ways. Therefore, the number of ways the frog can jump 5 times is $\left(2^{3}-2\right) \times 2^{2}=24$.
From (1) and (2), we know that the frog has a total of $2+24=26$ different ways to jump.