Proving binomial sum equals $0$ My hypothesis is that, when $n \equiv 0 \mod 6$, then
$$\sum_{k=0}^{n/3-1} \Bigg( \binom{n-2}{3k+1} 2^{3k+1}-\binom{n-2}{3k-1}2^{3k-1} \Bigg) = 0, \quad \binom{n-2}{3k-1} = 0 \text{ when } k=0$$
But I get stuck at finding a proof. I have tried induction, but that does not seem to work because the binomials change when increasing $n$. I have tried writing
$$\binom{n-2}{3k-1} = \binom{n-2}{3k+1} \frac{3k(3k+1)}{(n-3k-2)(n-3k-1)}$$
but I got stuck again. I tried writing out the terms for small $n$, making 2 summations instead of $1$ but I could not figure out the answer. Perhaps there is a counterexample but with such big numbers it is hard to check.

Hint: Let $$f_n(x)=\sum_{j=0}^n\binom{n}{j}x^j=(1+x)^n$$ and consider $f_n(2e^{2\pi i/3})-f_n(2e^{-2\pi i/3})$.
