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Hello everyone how can I calculate this expression

$\binom {2016}0 - \binom {2016}3 + \binom {2016}6 - \binom {2016}9 + ... +\binom {2016}{2016}$?

I tried to mark $\omega = \frac{\sqrt{3}i-1}{2}$ and $\omega^3 = 1$,

But I don't know how to continue.

eee
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    Try to expand $(1+w^k)^{2016}$ for $0\leq k\leq 5$, where $w$ is a primitive 6-th root of unity. – richrow Jun 25 '20 at 12:32
  • See https://math.stackexchange.com/questions/142260/sum-of-every-kth-binomial-coefficient – lhf Jun 25 '20 at 12:50

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Hint: Instead, let $\omega$ be such that $\omega^6 = 1$ (so $\omega = \frac 12 + i\frac{\sqrt{3}}2$). Let $f(x) = (1 + x)^{2016}$. Consider the sum $$ f(\omega) + f(\omega^3) + f(\omega^5). $$

We can calculate the above quantity as $$ (1 + \omega)^{2016} + (1 + \omega^{3})^{2016} + (1 + \omega^5)^{2016} = \\ (1 + \omega)^{2016} + (1 + \omega^5)^{2016} + (1 - 1)^{2016} = \\ 2 \operatorname{Re}[(1 + \omega)^{2016}]. $$ Note that $$ 1 + \omega = \frac 32 + i \frac{\sqrt 3}{2} = \sqrt{3} \left[\frac{\sqrt{3}}{2} + \frac 12i\right] = \sqrt{3}\alpha, $$ where we note that $\alpha^{12} = 1$. Noting that $2016 = 12 \cdot 168$, we find that $$ (1 + \omega)^{2016} = 3^{2016/2}\alpha^{2016} = 3^{1008}. $$

Ben Grossmann
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