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 jumping sequences can the frog have from the start to the stop? $\qquad$ 
(Provided by the Problem Group)
5. 26 .

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 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 $2^{3}-2$ ways, and the number of ways for the last two jumps is $4^{2}$ ways. Therefore, the number of ways the frog can jump 5 times is $\left(2^{3}-2\right) \cdot 2^{2}=24$ ways.
From (1) and (2), we know that the frog has $2+24=26$ different ways to jump.