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?
 A: Embed $a_n$ in a projective coordinate $[a_n,1]$ for each $n$. Then the given recursive relation may be written as
$$\left[\begin{array}{c}a_{n+1}\\ 1\end{array}\right]=\left[\begin{array}{cc}3&-1\\ -1&3\end{array}\right]\left[\begin{array}{c}a_n\\ 1\end{array}\right].$$
It follows by recursion that $$\left[\begin{array}{c}a_n\\ 1\end{array}\right]=A^{n-1}\left[\begin{array}{c}a_1\\ 1\end{array}\right],n\geq 1~\quad (1)$$ where $$A=\left[\begin{array}{cc}3&-1\\ -1&3\end{array}\right].$$
Diagonalize $A$ to get $$ \frac 12\left[\begin{array}{cc}1&1\\ 1&-1\end{array}\right]A\left[\begin{array}{cc}1&1\\ 1&-1\end{array}\right]=\left[\begin{array}{cc}2&0\\ 0&4\end{array}\right]$$ $$\Rightarrow A^{n-1}=\left[\begin{array}{cc}1&1\\ 1&-1\end{array}\right]\left[\begin{array}{cc}2^{n-1}&0\\ 0&2^{2n-2}\end{array}\right]\frac 12\left[\begin{array}{cc}1&1\\ 1&-1\end{array}\right]$$
$$=\left[\begin{array}{cc}2^{n-2}+2^{2n-3}&2^{n-2}-2^{2n-3}\\ 2^{n-2}-2^{2n-3}&2^{n-2}+2^{2n-3}\end{array}\right].$$
Substituting this back to (1), one gets that 
$$\left[\begin{array}{c}a_n\\ 1\end{array}\right]=\left[\begin{array}{cc}2^{n-2}+2^{2n-3}&2^{n-2}-2^{2n-3}\\ 2^{n-2}-2^{2n-3}&2^{n-2}+2^{2n-3}\end{array}\right]\left[\begin{array}{c}a_1\\ 1\end{array}\right]$$
$$\Leftrightarrow a_n=\frac{(2^{n-2}+2^{2n-3})a_1+(2^{n-2}-2^{2n-3})}{(2^{n-2}-2^{2n-3})a_1+(2^{n-2}+2^{2n-3})}=\frac{(2^{n-1}+1)a_1-2^{n-1}+1}{2^{n-1}+1-(2^{n-1}-1)a_1},$$ as required. QED
A: 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 $\alpha_{n}$
A: 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}
A: 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.
