5. If the expansion of $(1+x+x^2)^{10000}$ is $a_{0}+a_{1} x+a_{2} x^{2}+\cdots+a_{20000} x^{20000}$, then the value of $a_{0}+a_{3}+a_{6}+a_{9}+\cdots+a_{1998}$ is
(A) 3333
(B) 3666
(C) 3999
(D) 32001
(C)
5.【Analysis and Solution】Since the sum of every two consecutive coefficients in the expansion is required, we think of the unit roots of 1 and use the special value method.
Let $\omega=-\frac{1}{2}+\frac{\sqrt{3}}{2}$, then $\omega^{3}=1, \omega^{2}+\omega+1=0$.
Let $x=1$, we get
$$
3^{1000}+a_{0}+a_{1}+a_{2}+a_{3}+\cdots+a_{2000} \text {; }
$$

Let $x=\omega$, we get
$$
0=a_{0}+a_{1} \omega+a_{2} \omega^{2}+\cdots+a_{2000} \omega^{2000} \text {; }
$$

Let $x=\omega^{2}$, we get
$$
0=a_{0}+a_{1} \omega^{2}+a_{2} \omega^{4}+a_{3} \omega^{6}+\cdots+a_{2000} \omega^{4000} \text {; }
$$
(1) + (2) + (3) gives
$$
\begin{array}{l}
3^{1000}=3\left(a_{0}+a_{3}+a_{6}+\cdots+a_{1998}\right) \\
\therefore a_{0}+a_{3}+a_{6}+\cdots+a_{1998}=3^{9999}, \text { choose C. }
\end{array}
$$