<s>[INST] Prove $a_n=\frac{(2^{n-1}+1)a_1-2^{n-1}+1}{2^{n-1}+1-(2^{n-1}-1)a_1}$ for the recursive sequence $a_{n+1}=\frac{3a_n-1}{3-a_n}$ 
Prove the statement:
  $$a_n=\frac{(2^{n-1}+1)a_1-2^{n-1}+1}{2^{n-1}+1-(2^{n-1}-1)a_1}$$
  for the given recursive sequence: 
  $$a_{n+1}=\frac{3a_n-1}{3-a_n}. $$

My attempt:
Proof by induction:

(1) Base: $\tau(1)$. 
  $$
\begin{align}
a_2 &=\frac{(2+1)a_1-2+1}{2+1-(2-1)a_1} \\
&=\frac{3a_1-1}{3-a_1}.
\end{align}
$$
(2) Assumption: Let 
  $$ a_n=\frac{(2^{n-1}+1)a_1-2^{n-1}+1}{2^{n-1}+1-(2^{n-1}-1)a_1} $$ 
  hold for some $n\in\mathbb N$. 
(3)  Step $\tau(n+1)$:
  $$ 
\begin{align}
a_n &= \frac{3\cdot\frac{(2^{n-1}+1)a_1-2^{n-1}+1}{2^{n-1}+1-(2^{n-1}-1)a_1}-1}{3-\frac{(2^{n-1}+1)a_1-2^{n-1}+1}{2^{n-1}+1-(2^{n-1}-1)a_1}} \\
&= \frac{3\cdot2^{n-1}a_1+3a_1-3\cdot2^{n-1}+3-2^{n-1}-1+2^{n-1}a_1-a_1}{3\cdot 2^{n-1}+3-3\cdot 2^{n-1}a_1+3a_1-2^{n-1}a_1-a_1+2^{n-1}-1} \\
&= \frac{(4\cdot 2^{n-1}+2)a_1-4\cdot 2^{n-1}+2}{4\cdot 2^{n-1}+2-(4\cdot 2^{n-1}+2)a_1} \\
&= \frac{(2^n+1)a_1-2^n+1}{2^n+1-(2^n+1)a_1}
\end{align}
$$

Is this correct and is there a more efficient method than induction?
 [/INST] Let
$$a_n=\frac{x_n}{y_n}$$
Then
$$\frac{x_{n+1}}{y_{n+1}}=\frac{3x_n-y_n}{3y_n-x_n}$$
Assume
\begin{align}
&x_{n+1}=3x_n-y_n&
&y_{n+1}=3y_n-x_n
\end{align}
Then $y_n=3x_n-x_{n+1}$ from which $3x_{n+1}-x_{n+2}=3(3x_n-x_{n+1})-x_n$ which gives
$$x_{n+2}-6x_{n+1}+8x_n=0$$
This linear recurrence has general solution
$$x_n=2^{n-1}(4x_0-x_1)+2^{2n-1}(x_1-2x_0)$$
By symmetry, we have
$$y_n=2^{n-1}(4y_0-y_1)+2^{2n-1}(y_1-2y_0)$$
hence
\begin{align}
a_n
&=\frac{2^{n-1}(4x_0-x_1)+2^{2n-1}(x_1-2x_0)}{2^{n-1}(4y_0-y_1)+2^{2n-1}(y_1-2y_0)}\\
&=\frac{(4x_0-x_1)+2^{n}(x_1-2x_0)}{(4y_0-y_1)+2^{n}(y_1-2y_0)}\\
&=\frac{(2^n-1)x_1-4x_0(1-2^{n-1})}{(2^n-1)y_1-4y_0(1-2^{n-1})}\\
\end{align}
</s>