Given $a_{n} = 3a_{n-1}-2a_{n-2}$, $a_0=0$, $a_1 = 2$. Show $a_n = 2(2^n-1)$ by induction This is my problem:

The recurrence relation is \begin{aligned} a_{n} = 3a_{n-1}-2a_{n-2} \end{aligned}
This is given that $a_{0}=0, a_{1}=2$.

From the above information I calculated
\begin{aligned} a_{2} &amp;=3(a_{2 - 1}) - 2(a_{2 - 2}) = 3(a_{1}) - 2(a_{0}) = 3(2) - 2(0) = 6  \\
a_{3} &amp;=3(a_{3 - 1}) - 2(a_{3 - 2}) = 3(a_{2}) - 2(a_{1}) = 3(6) - 2(2) = 14 \\
 a_{4} &amp;=3(a_{4 - 1}) - 2(a_{4 - 2}) = 3(a_{3}) - 2(a_{2}) = 3(14) - 2(6) = 30 \end{aligned}
I have been trying to solve this recurrence relation for quite some time and have come up with the solution
\begin{aligned} a_{n} = 2 * (2^{n} - 1) \end{aligned}
I am having trouble proving this solution by induction.
My Attempt:
Base Cases:
$$a_0 = 0 = 2(2^0-1)\\a_1 = 2 = 2(2^1-1)$$
Inductive Hypothesis:
Assume that $a_k = 2(2^k-1)$ and $a_{k-1} = 2(2^{k-1}-1)$
Inductive Step:
$$a_{k+1} = 3a_{(k+1)-1}-2a_{(k+1)-2} = 3a_{k}-2a_{k-1}$$
By the inductive hypothesis  $a_k = 2(2^k-1)$ and $a_{k-1} = 2(2^{k-1}-1)$,
$$a_{k+1}= 3 ( 2 (2^k-1) - 2 (2(2^{k-1}-1)).$$
From here it is just simplifying but I cannot get it to simplify to the correct expression.
Any ideas?
Assume it holds for $n=k$ (so it also holds for $n&lt;k$), then for the induction step
\begin{align}
a_{k+1}&amp;=3a_{k}-2a_{k-1}\\
&amp;=3(2(2^{k}-1))-2(2(2^{k-1}-1))\\
&amp;=3(2^{k+1}-2)-2(2^{k}-2)\\
&amp;=3\cdot2^{k+1}-6-2^{k+1}+4\\
&amp;=2\cdot2^{k+1}-2\\
&amp;=2(2^{k+1}-1)
\end{align}
as required.
The claim that you want to prove by induction is
$$a_0=0\land a_1=2\land\forall n&gt;2:a_n=3a_{n-1}-2a_{n-2}\iff\forall n:a_n=2(2^n-1).$$
There is in fact no induction here, and you can use a direct proof,
$$2(2^n-1)=3\cdot2(2^{n-1}-1)-2\cdot2(2^{n-2}-1)=3\cdot2^n-6-2^n+4,$$ which is true.
You could force induction by stating that if the property holds for all $n\le m$, then it holds for all $n\le m+1$, but this is a little overkill.