9. (16 points) Given the sequence $\left\{a_{n}\right\}$ with the general term formula
$$
a_{n}=\frac{1}{\sqrt{5}}\left(\left(\frac{1+\sqrt{5}}{2}\right)^{n}-\left(\frac{1-\sqrt{5}}{2}\right)^{n}\right)\left(n \in \mathbf{Z}_{+}\right) \text {. }
$$

Let $S_{n}=\mathrm{C}_{n}^{1} a_{1}+\mathrm{C}_{n}^{2} a_{2}+\cdots+\mathrm{C}_{n}^{n} a_{n}$. Find all positive integers $n$ such that $7 \mid S_{n}$.
$$
\begin{aligned}
= & \frac{1}{\sqrt{5}}\left(\left((1+\alpha)^{n+1}-(1+\beta)^{n+1}\right) \cdot\right. \\
& (1+\alpha+1+\beta)- \\
& \left.(1+\alpha)(1+\beta)\left((1+\alpha)^{n}-(1+\beta)^{n}\right)\right) \\
= & 3 S_{n+1}-S_{n} .
\end{aligned}
$$

Given $S_{1}=\mathrm{C}_{1}^{1} a_{1}, S_{2}=\mathrm{C}_{2}^{1} a_{1}+\mathrm{C}_{2}^{2} a_{2}=3$, and $S_{n+2} = 3 S_{n+1} - S_{n}$, we can calculate the remainders of $\{S_{n}\}$ when divided by 7 as follows:
$$
1,3,1,0,6,4,6,0,1,3, \cdots
$$

This is a sequence with a period of 8.
$$
\text{Therefore, } 7\left|S_{n} \Leftrightarrow 4\right| n \text{, i.e., } n=4 k(k=1,2, \cdots) \text{. }
$$

Two, 9. Let $\alpha=\frac{1+\sqrt{5}}{2}, \beta=\frac{1-\sqrt{5}}{2}$.
Then $S_{n}=\frac{1}{\sqrt{5}} \sum_{i=1}^{n} \mathrm{C}_{n}^{i}\left(\alpha^{i}-\beta^{i}\right)$
$$
=\frac{1}{\sqrt{5}}\left((1+\alpha)^{n}-(1+\beta)^{n}\right) \text{. }
$$

Thus, $S_{n+2}$
$$
\begin{aligned}
= & \frac{1}{\sqrt{5}}\left(\left((1+\alpha)^{n+2}-(1+\beta)^{n+2}\right) \cdot\right. \\
& (1+\alpha+1+\beta)- \\
& \left.(1+\alpha)(1+\beta)\left((1+\alpha)^{n+1}-(1+\beta)^{n+1}\right)\right) \\
= & 3 S_{n+1}-S_{n} .
\end{aligned}
$$