Proving binomial sum equals $0$

Question

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.

Answer 1

Let $\omega=-\frac12+i\frac{\sqrt 3}2$, a primitive third root of unity. Then $$ (1+2\omega)^{n-2}= \sum_{r=0}^{n-2}{n-2\choose r}(2\omega)^r.$$ The imaginary part of the $r$th summand on the right is $0$ if $r=3k$, it is ${n-2\choose 2k+1}2^{3k+1}\cdot\frac{\sqrt 3}2$ if $r=3k+1$, and $-{n-2\choose 2k-1}2^{3k-1}\cdot\frac{\sqrt 3}2$ if $r=3k-1$. Hence your claim is equivalent to $$ \operatorname{Im}\left((1+2\omega)^{n-2}\right)=0.$$ And indeed for $n=6m$, $$(1+2\omega)^{n-2}=(i\sqrt 3)^{6m-2}=(-3)^{3m-1}\in\Bbb R.$$

Answer 2

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