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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.

I have asked this on AOPS forum too but people say it cannot be solved without using complex numbers. Please help me!

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If you assume the case $n:=3m$, $m \in \mathbb{N}_{0}$, you can write

$\sum_{r = 0}^m {3m \choose 3r}$

Wolframalpha gives for this:

$\frac{1}{3}\left(2(-1)^{m} + 8^{m}\right)$

Wolframalpha Calculation

I'm not completly able to reproduce this result, but this is maybe a good starting point.

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