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?

This is my solution
Let 
$$ \alpha_n = V_n + \beta    $$
$$a_{n+1} = V_{n+1} + \beta$$
then
$$V_{n+1} = \frac{3V_n + V_n\beta + (\beta^2 - 1)}{3-V_n-\beta} $$
At here we choose $\beta = 1$ as result from $\beta^2 - 1 = 0$
So we got $$V_{n+1} = \frac{4V_n}{2-V_n}$$
Continue we let
$$V_n = \frac{1}{u_n}$$
then
$$\frac{4\frac{1}{u_n}}{2-\frac{1}{u_n}} = \frac{1}{u_{n+1}}$$
$$ <=> \frac{1}{2}u_n - \frac{1}{4} = u_{n+1}$$
From here we need to find $u_1 $based on $v_1$ ,$\beta$ and $a_1$ then find $u_n$,$V_n$. Finally we will find out $\alpha_{n}$
