<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] For each $n = 1, 2, 3, \ldots$, we have
$$
\begin{align}
a_{n+1} &= \frac{3a_n -1 }{3 - a_n } \\
&= \frac{ 3a_n - 9 + 8 }{ 3 - a_n } \\
&= -3 + \frac{8}{ 3 - a_n }.
\end{align}
$$
Thus we obtain
$$ a_2 = - 3 + \frac{8}{3-a_1} = \frac{3a_1 - 1}{3 - a_1}. $$
$$ a_3 = -3 + \frac{ 8 }{ 3 - a_2 } = -3 + \frac{8}{3 - \frac{3a_1 - 1}{3 - a_1} } = ... $$
This approach might make it easier for you to proceed.
</s>