Three. (20 points) Given $A=\{1,2,3\}, x_{i} \in A$, $i=1,2, \cdots, 2006$. Let
$$
\begin{array}{l}
a_{1}=x_{1}+x_{2}+\cdots+x_{2006}, \\
a_{2}=x_{1}^{2}+x_{2}^{2}+\cdots+x_{2006}^{2}, \\
a_{n}=x_{1}^{n}+x_{2}^{n}+\cdots+x_{2000}^{n}\left(n \in \mathbf{N}_{+}\right) .
\end{array}
$$

Try to express $a_{n}$ in terms of $a_{1}$ and $a_{2}$.
Three, let $x_{i}(i=1,2, \cdots, 2006)$ take $1,2,3$ respectively $s, t, r$ times. Then
$$
\left\{\begin{array}{l}
s+t+r=2006, \\
a_{1}=s+2 t+3 r=2006+t+2 r, \\
a_{2}=s+4 t+9 r=2006+3 t+8 r .
\end{array}\right.
$$

Solving, we get $t=4 a_{1}-a_{2}-3 \times 2006$,
$$
r=2006-\frac{3 a_{1}-a_{2}}{2} \text {. }
$$

Thus, $a_{n}=s+2^{n} t+3^{n} r$
$$
\begin{aligned}
= & 2006+\left(2^{n}-1\right) t+\left(3^{n}-1\right) r \\
= & 2006+\left(2^{n}-1\right)\left(4 a_{1}-a_{2}-3 \times 2006\right)+ \\
& \left(3^{n}-1\right)\left(2006-\frac{3 a_{1}-a_{2}}{2}\right) \\
= & \left(3^{n}-3 \times 2^{n}+3\right) 2006+ \\
& \left(2^{n+2}-\frac{1}{2} \times 3^{n+1}-\frac{5}{2}\right) a_{1}+ \\
& \left(\frac{1}{2} \times 3^{n}-2^{n}+\frac{1}{2}\right) a_{2} .
\end{aligned}
$$