8.34 The sequence of real numbers $a_{0}, a_{1}, a_{2}, \cdots, a_{n}, \cdots$ satisfies the following equation: $a_{0}=a$, where $a$ is a real number,
$$
a_{n}=\frac{a_{n-1} \sqrt{3}+1}{\sqrt{3}-a_{n-1}}, n \in N .
$$

Find $a_{1994}$.
［Solution］After calculation, we can obtain
$$
\begin{aligned}
a_{1} & =\frac{a \sqrt{3}+1}{\sqrt{3}-a}, \\
a_{2} & =\frac{a_{1} \sqrt{3}+1}{\sqrt{3}-a_{1}}=\frac{(a \sqrt{3}+1) \sqrt{3}+(\sqrt{3}-a)}{\sqrt{3}(\sqrt{3}-a)-(a \sqrt{3}+1)} \\
& =\frac{a+\sqrt{3}}{1-a \sqrt{3}}, \\
a_{3} & =\frac{a_{2} \sqrt{3}+1}{\sqrt{3}-a_{2}}=\frac{(a+\sqrt{3}) \sqrt{3}+(1-a \sqrt{3})}{\sqrt{3}(1-a \sqrt{3})-(a+\sqrt{3})} \\
& =-\frac{1}{a} .
\end{aligned}
$$

From this, we get
$$
a_{6}=-\frac{1}{-\frac{1}{a}}=a .
$$

It is evident that the sequence $a_{0}, a_{1}, a_{2}, \cdots$ is a periodic sequence with a period of 6.
Since $1994=6 \times 332+2$, we have
$$
a_{1994}=a_{2}=\frac{a+\sqrt{3}}{1-a \sqrt{3}} .
$$