6. Let $(x+1)^{2002}=a_{0}+a_{1} x+a_{2} x^{2}+\cdots+a_{2002} x^{2002}$, then $a_{1}+a_{4}+$ $a_{7}+\cdots+a_{2002}=(\quad)$.
A. $\frac{2^{2002}-1}{3}$
B. $\frac{2^{2002}+2}{3}$
C. $2^{2002}$
D. 0
6. A

Let $S_{0}=a_{0}+a_{3}+\cdots, S_{1}=a_{1}+a_{4}+\cdots, S_{2}=a_{2}+a_{5}+\cdots$,
Let $f(x)=(x+1)^{2002}$, and $W=e^{\frac{2 \pi}{3}}$,
Then $S_{0}+S_{1}+S_{2}=f(1)=2^{2002}$;
$$
\begin{array}{l}
S_{0}+W S_{1}+W^{2} S_{2}=f(W)=\left(-W^{2}\right)^{2002}=W^{4004}=W^{2} ; \\
S_{0}+W^{2} S_{1}+W S_{2}=f\left(W^{2}\right)=(-W)^{2002}=W^{2002}=W ;
\end{array}
$$

By (1) + (2) $\times W^{2}+$ (3) $\times W$, we get
$$
\begin{aligned}
& \left(1+W^{2}+W\right) \cdot S_{0}+3 S_{1}+\left(1+W+W^{2}\right) S_{2}=2^{2002}+W^{4}+W^{2} \\
\Rightarrow & 3 \cdot S_{1}=2^{2002}-1
\end{aligned}
$$

Therefore, $S_{1}=\frac{2^{2002}-1}{3}$.