6. The number of 2007-digit decimal numbers $\overline{2 a_{1} a_{2} a_{3} \cdots a_{2006}}$ with an odd number of 9s among the digits $a_{1}, a_{2}, a_{3}, \cdots, a_{2006}$ is
A. $\frac{1}{2}\left(10^{2006}+8^{2006}\right)$
B. $\frac{1}{2}\left(10^{2006}-8^{2006}\right)$
C. $10^{2006}+8^{2006}$
D. $10^{2006}-8^{2006}$
The number of $\overline{2 a_{1} a_{2} a_{3} \cdots a_{2006}}$ with an odd number of 9s is $C_{2006}^{1} 9^{2005}+C_{2006}^{3} 9^{2003}+\cdots+C_{2006}^{2005} 9$.
Let $A=C_{2006}^{0} 9^{2006}+C_{2006}^{2} 9^{2004}+\cdots+C_{2006}^{2004} 9^{2}+C_{2006}^{2006}$,
$$
B=C_{2006}^{1} 9^{2005}+C_{2006}^{3} 9^{2003}+\cdots+C_{2006}^{2005} 9 \text {, }
$$
$\left\{\begin{array}{l}A+B=(9+1)^{2006}=10^{2006}, \\ A-B=(9-1)^{2006}=8^{2006}\end{array} \Rightarrow B=\frac{1}{2}\left(10^{2006}-8^{2006}\right)\right.$. Therefore, the answer is $B$.