6 digits $a_{1}, a_{2}, a_{3}, \cdots, a_{2} 006$ in which there are an odd number of 9s in the 2007-digit decimal number $2 a_{1} a_{2} a_{3} \cdots a_{2006}$ is ( ).
(A) $\frac{1}{2}\left(10^{2006}+8^{2006}\right)$
(B) $\frac{1}{2}\left(10^{2006}-8^{2005}\right)$
(C) $10^{2006}+8^{2006}$
(D) $10^{2006}-8^{2006}$
Solve for the number of decimal numbers that have an odd number of 9s:
$$
A=\mathrm{C}_{2006}^{1} 9^{2005}+\mathrm{C}_{2006}^{3} 9^{2003}+\cdots+\mathrm{C}_{2006}^{2005} 9 \text {. }
$$

Also, since
$$
(9+1)^{2006}=\sum_{k=0}^{2006} C_{2006}^{k} 9^{2006-k}
$$

and
$$
(9-1)^{2006}=\sum_{k=0}^{2006} C_{2006}^{k}(-1)^{k} 9^{2006-k},
$$

we get
$$
\begin{aligned}
A & =\mathrm{C}_{2006}^{1} 9^{2005}+\mathrm{C}_{2006}^{3} 9^{2003}+\cdots+\mathrm{C}_{2006}^{2005} 9 \\
& =\frac{1}{2}\left(10^{2006}-8^{2006}\right) .
\end{aligned}
$$

Therefore, the answer is B.