<s>[INST] Find the closed formula for the following recusive sequence. Find the closed formula for the recursive sequence defined by $a_0 = 4$, $a_1 = 12$, $a_n = 6a_{n-1}-a_{n-2}$ for $n>1$. This question is stumping me. I don't know any methods besides guess and check.
Current progress:
The first five terms are $4, 12, 68, 396, 2308$.
Each term is divisible by four, yielding $1, 3, 17, 99, 577$
Each of these is off of a perfect square by one or negative one $0-1, 4-3, 16-17, 100-99, 576-577$ = $-1,1,-1,1,-1$
Other than that I'm lost. Any help would be appreciated.
 [/INST] 
Using the following fact,
$$\begin{cases}a_{n}=ba_{n-1}+ca_{n-2},\\a_{0}=c,\quad a_{1}=d \end{cases}$$ then $$\begin{cases} a_{n}=\alpha r_{1}^{2}+\beta r_{2}^{2}, \quad \text{if} \quad r_{1}\not=r_{2},\quad r^{2}-br-c=0,\\a_{n}=br_{1}^{n}+dnr_{2}^{n},\quad \text{if}\quad r_{1}=r_{2},\quad r^{2}-br-c=0,\end{cases}$$
where $\alpha$ and $\beta$ can be found using the initial condition $a_{0}=c$ and $a_{1}=d$ by substitution in $a_{n}$.

Then, we have the recursive sequence with initial condition, $$\begin{cases}a_{n}=6a_{n-1}+(-1)\cdot a_{n-2},\\a_{0}=4,\quad a_{1}=12\end{cases}$$
Since $$r^{2}-6r-(-1)=0,\quad \text{then}\quad r=\frac{6\pm \sqrt{6^{2}-4}}{2},\quad \text{then}\quad r=3\pm \sqrt{8}, \quad \text{then}\quad r_{1}\not=r_{2}.  $$
Then, the general solution is given by
$$a_{n}=\alpha\left(3+\sqrt{8}\right)^{n}+\beta\left(3-\sqrt{8}\right)^{n}.$$
Setting $a_{0}=4$ and $a_{1}=12$ we have
$$\begin{cases}4=\alpha\left(3+\sqrt{8}\right)^{0}+\beta\left(3-\sqrt{8}\right)^{0},\quad n=0,\\ 12=\alpha\left(3+\sqrt{8}\right)^{1}+\beta\left(3-\sqrt{8}\right)^{1},\quad n=1.\end{cases},\quad \text{then}\quad \begin{cases}\alpha=2,\\ \beta=2.\end{cases}$$
Therefore, the particular solution for the recursive sequence is given by
$$\color{blue}{\boxed{a_{n}=2\left(3+\sqrt{8}\right)^{n}+2\left(3-\sqrt{8}\right)^{n},\quad n=0,1,\ldots}}$$
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