3.2.6 * If the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=5$, and $a_{n+1}=\frac{3 a_{n}-1}{-a_{n}+3}, n=1,2, \cdots$, find the general term $a_{n}$.
This is a problem involving a fractional recursive sequence. Let $b_{n}=a_{n}-1$, then
$$
b_{n+1}+1=\frac{3\left(b_{n}+1\right)-1}{-\left(b_{n}+1\right)+3}=\frac{3 b_{n}+2}{-b_{n}+2},
$$

so $b_{n+1}=\frac{4 b_{n}}{-b_{n}+2}, b_{1}=4$, it is easy to see that $b_{n} \neq 0$, thus $\frac{1}{b_{n+1}}=\frac{1}{2} \cdot \frac{1}{b_{n}}-\frac{1}{4}$. Therefore, it is easy to find the general term of the sequence $\left\{\frac{1}{b_{n}}\right\}$ as $\frac{1}{b_{n}}=\frac{3-2^{n}}{2^{n+1}}$, so $a_{n}=\frac{3+2^{n}}{3-2^{n}}$.