6. Three people, A, B, and C, practice passing a ball, making a total of $n$ passes. The ball starts from A and the $n$-th pass is still to A, with a total of $a_{n}$ different ways to pass the ball. Let the sequence of passing methods be $\left\{a_{n}\right\}$. Then the general term formula for the sequence $\left\{a_{n}\right\}$ is ( ).
(A) $a_{n}=\frac{1}{3} \times 2^{n}+\frac{2}{3}$
(B) $a_{n}=\frac{1}{3} \times 2^{n}-\frac{2}{3}$
(C) $a_{n}=\frac{1}{3} \times 2^{n}+(-1)^{n} \frac{2}{3}$
(D) Cannot be determined
6.C.

The number of ways the ball is passed to A on the $n$th pass is $a_{n}$. Let the number of ways the ball is not passed to A on the $n$th pass be $b_{n}$. For A, since there are 2 ways to pass the ball each time, there are $2^{n}$ ways for $n$ passes in total. Therefore, $a_{n}+b_{n}=2^{n}$.

Additionally, if the ball is passed to A on the $(n+1)$th pass, then the ball must not have been passed to A on the $n$th pass, so $a_{n+1}=b_{n}$.
Thus, $a_{n+1}+a_{n}=2^{n}$.
Let $c_{n}=\frac{a_{n}}{2^{n}}$, then $c_{n+1}+\frac{1}{2} c_{n}=\frac{1}{2}$.
Therefore, $c_{n+1}-\frac{1}{3}=-\frac{1}{2}\left(c_{n}-\frac{1}{3}\right)$.
Since $a_{1}=0, c_{1}=0, c_{1}-\frac{1}{3}=-\frac{1}{3}$, the sequence $\left\{c_{n}-\frac{1}{3}\right\}$ is a geometric sequence with the first term $-\frac{1}{3}$ and common ratio $-\frac{1}{2}$. Thus,
$$
c_{n}=\frac{1}{3}+\left(-\frac{1}{3}\right)\left(-\frac{1}{2}\right)^{n-1} .
$$

Therefore, $a_{n}=\frac{1}{3} \times 2^{n}+(-1)^{n} \times \frac{2}{3}$.