Solving difference equations 
Given $$\alpha\{v(a + 1) - v(1) \} = \beta v(a)$$
Deduce $$v(a) = \frac{1-(\beta/\alpha)^m}{1-(\beta/\alpha)} v(1),$$ and derive $$v(a) = \frac{1-(\beta/\alpha)^a}{1-(\beta/\alpha)^{m+n}}.$$

I am new to difference equations, but I am following wiki.
$$v(a+1) = \frac{\beta}{\alpha}v(a) + v(1),$$ which has steady state value $$y^*= \frac{v(1)}{1-\frac{\beta}{\alpha}}.$$
Moving on from here I find the solution to the homogeneous problem as
$$x(a) = \bigg(\frac{\beta}{\alpha}\bigg)^av(1).$$
This is not correct. What am I doing wrong? Even if I find the first part I can not see how to progress to the second part ( the derivation part).
 A: There seems to be some information missing regarding $m,n$. But we can make some progress without its knowledge. We start with the recurrence relation
\begin{align*}
\alpha\left(\nu(a+1)-\nu(1)\right)=\beta \nu(a)\qquad\qquad a\geq 1, \alpha,\beta \ne 0
\end{align*}
We can write the recurrence relation as
\begin{align*}
\color{blue}{\nu(a+1)-\frac{\beta}{\alpha}\nu(a)=\nu(1)\qquad\qquad a\geq 1}\tag{1}
\end{align*}

We use (1) to write down a chain of recurrence equations, each equation multiplied with a factor so that we can apply telescoping conveniently. We obtain
\begin{align*}
\color{blue}{\nu(a)}-\frac{\beta}{\alpha}\nu(a-1)&=\nu(1)\\
\frac{\beta}{\alpha}\,\nu(a-1)-\left(\frac{\beta}{\alpha}\right)^2\nu(a-2)&=\frac{\beta}{\alpha}\nu(1)\\
\left(\frac{\beta}{\alpha}\right)^2\nu(a-2)-\left(\frac{\beta}{\alpha}\right)^3\nu(a-3)&=\left(\frac{\beta}{\alpha}\right)^2\nu(1)\\
&\ \ \vdots\\
\left(\frac{\beta}{\alpha}\right)^{a-2}\nu(2)\color{blue}{-\left(\frac{\beta}{\alpha}\right)^{a-1}\nu(1)}&=\left(\frac{\beta}{\alpha}\right)^{a-2}\nu(1)\\
\end{align*}
Summing up these lines all but the blue marked terms at the left side cancel and we obtain using the finite geometric series formula
\begin{align*}
\nu(a)-\left(\frac{\beta}{\alpha}\right)^{a-1}\nu(1)&=\nu(1)\sum_{k=0}^{a-2}\left(\frac{\beta}{\alpha}\right)^{k}\\
\color{blue}{\nu(a)}&=\nu(1)\sum_{k=0}^{a-1}\left(\frac{\beta}{\alpha}\right)^{k}\\
&\color{blue}{=\nu(1)\frac{1-\left(\frac{\beta}{\alpha}\right)^{a}}{1-\frac{\beta}{\alpha}}}\tag{2}
\end{align*}

Setting
$
\nu(1)=\frac{1-\frac{\beta}{\alpha}}{1-\left(\frac{\beta}{\alpha}\right)^{m+n}}
$
we get from (2) OP's final expression
\begin{align*}
\color{blue}{\nu(a)=\frac{1-\left(\frac{\beta}{\alpha}\right)^{a}}{1-\left(\frac{\beta}{\alpha}\right)^{m+n}}}
\end{align*}
A: There is no need to invoke difference equations. To obtain $$v(a) = \frac{1-(\beta/\alpha)^a}{1-(\beta/\alpha)} v(1),$$
we just use induction and $$v(a + 1)  = \frac{\beta}{\alpha} v(a) + v(1).$$
Then we use the fact that $v(m+n) = 1$ (This was missing from my original post) to find $v(1)$ as $$\frac{1-\beta /\alpha}{1-(\beta /\alpha)^{m+n}}.$$
Plugging this in the equation for $v(a)$ at the top of this reply and we get the desired result.
