Solving Recurrence Relation of $T(n)=4T(n-2)+2$ Question

Solve Recurrence Relation of $T(n)=4T(n-2)+2$
Base case-: $T(1)=1,T(2)=2$

My Approach/solution
$$T(n)=4T(n-2)+2$$
$$T(n-2)=4T(n-4)+2 \tag{1}$$
$$T(n-4)=4T(n-6)+2 \tag{2}$$
Using $(1)$ and $(2)$ in my equation
$$\begin{align*}
T(n)&=4\cdot (4T(n-4)+2)+2\\
&=4^{2}\cdot T(n-2\cdot 2)+2\cdot 4^{1}+2\cdot 4^{0}\\
&=4^{2}\cdot(4T(n-6)+2)+2\cdot 4^{1}+2\cdot 4^{0}\\
&=4^{3}\cdot T(n-2\cdot 3)+2\cdot 4^{2}+2\cdot 4^{1}+2\cdot 4^{0}\\
\vdots \\
&=4^{k}\cdot T(n-2\cdot k)+2\cdot 4^{k-1}+...+2\cdot 4^{2}+2\cdot 4^{1}+2\cdot 4^{0}
\end{align*}$$
Substituing  $T(n-2\cdot k)$ by $2$, i.e $T(2)=2$
$$n-2\cdot k=2 \Rightarrow k=\frac{n-2}{2}$$
So our equation will look like
$$\begin{align*}
T(n)&=2\cdot 4^{k}+2\cdot 4^{k-1}+...+2\cdot 4^{2}+2\cdot 4^{1}+2\cdot 4^{0}\\
T(n)&=2\cdot \left(4^{0}+4^{1}+4^{2}+...+4^{k-1}+4^{k}\right)\\
T(n)&=2\cdot \left(4^{0}\cdot \frac{(4^{k+1}-1)}{4-1}\right)
\end{align*}$$
$k=\frac{n-2}{2}$
$$\begin{align*}
T(n)&=2\cdot \left(\frac{(4^{k+1}-1)}{4-1}\right)\\
T(n)&=2\cdot \frac{2^{n}-1}{3}
\end{align*}$$
Is it correct? Also if it is correct, can anyone hint me another approach as it is bit lengthy.
Thanks!

i think the right solution is given by $$T(n)=\frac{1}{12} \left(9\ 2^n+(-1)^{n+1} 2^n-8\right)$$
at first solve the equation $$T(n)=4T(n-2)$$ with $$T(n)=q^n$$
