Simplify without using complex numbers.

Simplify $$\large\sum_{r=0}^{\left \lfloor \frac{n}{3} \right \rfloor}\binom {n}{3r}$$

I tried as much as I could; tried to apply induction, tried to approach combinatorally but failed. I could not resist to see the solution after trying it for all the day. But the answer used complex numbers(cube root of unity and de’moiver’s) and I didn’t find it elegent anyway. In short, I am asking you to simplify it without using complex numbers.

If you assume the case $$n:=3m$$, $$m \in \mathbb{N}_{0}$$, you can write
$$\sum_{r = 0}^m {3m \choose 3r}$$
$$\frac{1}{3}\left(2(-1)^{m} + 8^{m}\right)$$