Identity for this recursive relation with multiple terms I have a recursive relation algorithm which is defined as follows:
$$F_n = 3(F_{n-1} - F_{n-2}) + F_{n-3}$$
$$F_0 = 0$$
$$F_1 = 1$$
$$F_2 = 4$$
From calculating the first few values, I know this is equivalent to $F_n = n^2$, however I don't quite know how to derive that. I tried substituting and telescoping, but I couldn't find an appropriate pattern.
How would I do this?

Proof
We want to show that $F_n = n^2$ for all $n \geq 0$. This is clearly true for $n=0,1,2$ since $F_0=0, F_1=1, F_2=4$. For $F_3$, we get that $F_3 = 3(F_2-F_1)+F_0 = 3(4-1)+0 = 9 = 3^2$. So it is true for $n=3$.
Assume for some $k\geq 3$ that $F_i = i^2$ for all $i \leq k$. We want to show that $F_{k+1} = (k+1)^2$. By the recurrence relation, $F_{k+1} = 3(F_k-F_{k-1})+F_{k-2}$. By the inductive hypothesis, we know that $F_k = k^2, F_{k-1} = (k-1)^2, F_{k-2}=(k-2)^2$. So 
$\begin{align}
F_{k+1} &= 3(k^2-(k-1)^2)+(k-2)^2 \\
&= 3[k^2-(k^2-2k+1)]+(k^2-4k+4) \\
&= 3(2k-1)+k^2-4k+4 \\ 
&= 6k-3+k^2-4k+4 \\
&= k^2+2k+1 = (k+1)^2
\end{align}$
