2 . If $x_{n+1}=\frac{3 x_{n}+4}{2 x_{n}+1}, x_{1}=1$, find $x_{n}$.


(Solution: Fixed points $\alpha=-1,2, \frac{x_{n-1}+1}{x_{n+1}-2}$ $=-2(-5)^{n}$. Solving for $x_{n+1}=\frac{4(-5)^{n}+1}{2(-5)^{n}+1}$, i.e., $\left.x_{n}=\frac{4(-5)^{n} \cdot 1-1}{2(-5)^{n} \cdot 1+1}.\right)$