Think about Question 1 Let $x_{1}=3$, and $x_{n}=\frac{x_{n-1}^{2}+2}{2 x_{n-1}-1}(n=2,3, \cdots)$, find the general term formula of the sequence $\left\{x_{n}\right\}$.
Solution: The characteristic equation $x=\frac{x^{2}+2}{2 x-1}$, i.e., $x^{2}-x-2=0$, has two roots $\alpha=2, \beta=-1$, so $x_{n}+1=\frac{x_{n-1}^{2}+2 x_{n-1}+1}{2 x_{n-1}-1}, x_{n}-2=\frac{x_{n-1}^{2}-4 x_{n-1}+4}{2 x_{n-1}-1}$,
thus $\frac{x_{n}+1}{x_{n}-2}=\left(\frac{x_{n-1}+1}{x_{n-1}-2}\right)^{2}=\left[\left(\frac{x_{n-2}+1}{x_{n-2}-2}\right)^{2}\right]^{2}=\cdots=\left(\frac{x_{1}+1}{x_{1}-2}\right)^{2^{n-1}}=4^{2^{n-1}}$, hence $x_{n}=\frac{-1-2 \times 4^{2^{n-1}}}{1-4^{2^{n-1}}}=\frac{2 \times 2^{2^{n}}+1}{2^{2^{n}}-1}$.
If the sequence $\left\{a_{n}\right\}$ satisfies the recurrence relation $x_{n+2}=\frac{x_{n+1}^{2}+p \cdot q^{n}}{x_{n}}$, and $x_{1}=a, x_{2}=b(a, b$ are constants $)$, then we can eliminate $q^{n}$ to convert it into a second-order linear recurrence sequence.