8 A certain intelligence station has four different passwords $A, B, C, D$, and uses one of them each week, and each week it randomly selects one from the three passwords not used in the previous week with equal probability. If the first week uses password $A$, what is the probability that the 7th week also uses password $A$? (Express in simplest fraction form)

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$8 \frac{61}{243}$ Hint: Let $P_{k}$ represent the probability of using password type $A$ in the $k$-th week, then the probability of not using password type $A$ in the $k$-th week is $1-P_{k}$. Thus, we have
$$
P_{k+1}=\frac{1}{3}\left(1-P_{k}\right), k \in \mathbf{N}^{*},
$$

which means
$$
P_{k+1}-\frac{1}{4}=-\frac{1}{3}\left(P_{k}-\frac{1}{4}\right) \text {. }
$$

Given $P_{1}=1$, we know that $\left\{P_{k}-\frac{1}{4}\right\}$ is a geometric sequence with the first term $\frac{3}{4}$ and common ratio $-\frac{1}{3}$. Therefore, $P_{k}-\frac{1}{4}=\frac{3}{4}\left(-\frac{1}{3}\right)^{k-1}$, which means $P_{k}=\frac{3}{4}\left(-\frac{1}{3}\right)^{k-1}+\frac{1}{4}$. Hence, $P_{7}=\frac{61}{243}$.