<s>[INST] Calculating $\binom {2016}0 - \binom {2016}3 + \binom {2016}6 - \binom {2016}9 + ... +\binom {2016}{2016}$ 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.
 [/INST] 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
$$
(1 + \omega)^{2016} = 3^{2016/2}\alpha^{2016} = 3^{1008}.
$$
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