Problem 817

Let $x=\frac{x+3}{2 x-4}$. Then
$$2 x^{2}-5 x-3=0 \text {. }$$

Therefore, $x_{1}=-\frac{1}{2}, x_{2}=3$ are the two fixed points of $f(x)=\frac{x+3}{2 x-4}$. Hence
$$\begin{array}{l} a_{n+1}+\frac{1}{2}=\frac{a_{n}+3}{2 a_{n}-4}+\frac{1}{2}=\frac{2\left(a_{n}+\frac{1}{2}\right)}{2 a_{n}-4}, \\ a_{n+1}-3=\frac{a_{n}+3}{2 a_{n}-4}-3=\frac{-5\left(a_{n}-3\right)}{2 a_{n}-4} . \end{array}$$
(1) ÷ (2) gives
$$\frac{a_{n+1}+\frac{1}{2}}{a_{n+1}-3}=-\frac{2}{5} \cdot \frac{a_{n}+\frac{1}{2}}{a_{n}-3} .$$

Therefore, the sequence $\left\{\frac{a_{n}+\frac{1}{2}}{a_{n}-3}\right\}$ is a geometric sequence with the first term $\frac{a_{1}+\frac{1}{2}}{a_{1}-3}=-\frac{2}{5}$ and common ratio $-\frac{2}{5}$.
$$\begin{array}{l} \text { Hence } \frac{a_{n}+\frac{1}{2}}{a_{n}-3}=\left(\frac{a_{1}+\frac{1}{2}}{a_{1}-3}\right)\left(-\frac{2}{5}\right)^{n-1} \\ =\left(-\frac{2}{5}\right)^{n} . \end{array}$$

Therefore, $a_{n}=\frac{(-5)^{n}+3 \times 2^{n+1}}{2^{n+1}-2(-5)^{n}}\left(n \in \mathbf{N}_{+}\right)$.