Finding $a_n$ if $a_{n+1}=3a_n-2a_{n-1}$ with $a_1=2$ and $a_2=3$ Given $a_1=2,\; a_2=3,\; a_{n+1}=3a_n-2a_{n-1}$.

How can i prove that:  $a_n=2^{n-1}+1$

I Tried to isolate $a_n$ but it doesn't get me anywhere.

Thanks.
The general method to solve the recursion $$a_{n+1}=\color{green}{\bf 3}a_n-\color{green}{\bf 2}a_{n-1},$$ is to consider the characteristic equation $$r^2=\color{green}{\bf 3}r-\color{green}{\bf 2},$$ whose roots are $r=\color{red}{\bf1}$ and $r=\color{red}{\bf2}$ hence there exists $A$ and $B$ such that, for every $n$, $$a_n=A\cdot\color{red}{\bf2}^n+B\cdot\color{red}{\bf1}^n=A\cdot2^n+B.$$
The initial conditions yield $A$ and $B$.

In full generality, the characteristic equation of the recurrence $$a_{n+1}=\color{green}{\mathbf b_1}a_n+\color{green}{\mathbf b_2}a_{n-1}+\cdots+\color{green}{\mathbf b_k}a_{n-k+1},$$ is the polynomial identity $$r^k=\color{green}{\mathbf b_1}r^{k-1}+\color{green}{\mathbf b_2}r^{k-2}+\cdots+\color{green}{\mathbf b_k}.$$
$$\begin{align}
a_{n+1}&amp;=3a_n-2a_{n-1}\\
a_{n+1}-a_n&amp;=2(a_n-a_{n-1})\\
u_{n+1}&amp;=2u_n\end{align}$$
where $u_n=a_n-a_{n-1}$

Given $a_1=2, a_2=3$, we have

$$\begin{align}
u_2&amp;=a_2-a_1&amp;=1\\
u_3&amp;=a_3-a_2&amp;=2\cdot 1 = 2\\
u_4&amp;=a_4-a_3&amp;=2\cdot 2 = 4\\
\vdots\\
u_n&amp;=a_n-a_{n-1}&amp;=2\cdot 2^{n-3}=2^{n-2}\end{align}$$

Summing and telescoping gives 
$$\begin{align}\require{cancel}a_n-\cancelto{2}{a_1}&amp;=\sum_{r=1}^{n-1}2^{r-1}\\
&amp;=\frac{2^{n-1}-1}{2-1}\\
&amp;=2^{n-1}-1\\
a^n&amp;=2^{n-1}+1\end{align}$$