3. The sequence $a_{0}, a_{1}, a_{2}, \cdots$ satisfies
$$
a_{0}=\sqrt{3}, a_{n+1}=\left[a_{n}\right]+\frac{1}{\left\{a_{n}\right\}}\left(\left[a_{n}\right]\right. \text { and }
$$
$\left\{a_{n}\right\}$ represent the integer part and the fractional part of $a_{n}$, respectively). Then $a_{2006}=$ $\qquad$ :
$$
3.3009+\sqrt{3} \text {. }
$$

From the given, we have
$$
\begin{array}{l}
a_{0}=1+(\sqrt{3}-1), \\
a_{1}=1+\frac{1}{\sqrt{3}-1}=1+\frac{\sqrt{3}+1}{2}=2+\frac{\sqrt{3}-1}{2}, \\
a_{2}=2+\frac{2}{\sqrt{3}-1}=2+(\sqrt{3}+1)=4+(\sqrt{3}-1), \\
a_{3}=5+\frac{\sqrt{3}-1}{2}, \\
a_{4}=7+(\sqrt{3}-1), \\
\cdots \cdots
\end{array}
$$

It is easy to see that, $a_{2 k}=3 k+1+(\sqrt{3}-1)$,
$$
a_{2 k+1}=3 k+2+\frac{\sqrt{3}-1}{2} \text {. }
$$

Therefore, $a_{2000}=3009+\sqrt{3}$.