9. The sequence $\left\{a_{n}\right\}$ satisfies: $a_{1}=a_{2}=a_{3}=1$. Let $b_{n}=a_{n}+a_{n+1}+a_{n+2}\left(n \in \mathbf{N}^{*}\right)$. If the sequence $\left\{b_{n}\right\}$ is a geometric sequence with a common ratio of 3, find the value of $a_{100}$.
$$
\begin{array}{l}
b_{1}=3 \Rightarrow b_{n}=3^{n}, \text { i.e., } a_{n+2}+a_{n+1}+a_{n}=3^{n}, a_{n+3}+a_{n+2}+a_{n+1}=3^{n+1} \\
\Rightarrow a_{n+3}-a_{n}=2 \cdot 3^{n}. \text { Thus, } a_{4}-a_{1}=2 \cdot 3, a_{7}-a_{4}=2 \cdot 3^{4}, a_{10}-a_{7}=2 \cdot 3^{7}, \cdots, \\
a_{100}-a_{97}=2 \cdot 3^{97} \text { for a total of } 33 \text { terms, } \\
\text { so } a_{100}=\frac{6\left(1-3^{99}\right)}{1-3^{3}}+1=\frac{3^{100}+10}{13} .
\end{array}
$$