5. Let $A B C D E F$ be a regular hexagon. A frog starts at vertex $A$ and can randomly jump to one of the two adjacent vertices each time. 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 possible jumping sequences are there from the start to the stop? $\qquad$
5. As shown in the figure, it is obvious that the frog cannot reach point $D$ after jumping 1, 2, or 4 times. Therefore, the frog's jumping methods are only of the following two scenarios:
（1）The frog reaches point $D$ after 3 jumps, with 2 ways to do so;
（2）The frog stops after a total of 5 jumps. In this case, the number of ways for the first 3 jumps (which definitely do not reach point $D$) is $2^{3}-2$, and the number of ways for the last two jumps is $2^{2}$. Therefore, the number of ways the frog can jump a total of 5 times is $\left(2^{3}-2\right) \cdot 2^{2}=24$.
From (1) and (2), we know that the frog has a total of $2+24=26$ different ways to jump.