How to prove that $$1+\binom{n}{3} + \binom{n}{6}+\binom{n}{9}+ \dots = \frac{1}{3} \Big( 2^n + 2 \cos \frac{\pi n}{3} \Big)$$ My attempt: we know that $2^n=\binom{n}{0}+\binom{n}{1}+\binom{n}{2}+\dots+\binom{n}{n}$ and from $(1+i)^n=\cos \frac{\pi n}{4}+i \cos \frac{\pi n}{4}=\sum_{k=0}^{n} \binom{n}{k} i^k$ we have $\binom{n}{0}-\binom{n}{2}+\binom{n}{4}-\dots=2^{\frac{n}{2}} \cos \frac{\pi n}{4}$ and $\binom{n}{1}-\binom{n}{3}+\binom{n}{5}-\dots=2^{\frac{n}{2}} \sin \frac{\pi n}{4}$. I tried to add and subtract somehow these identities in order to get the sum of $\binom{n}{3k}$, but this is bad approach because there is no $\cos\frac{\pi n}{3}$. I don't know how $\cos \frac{\pi n}{3}$ must appears. If we consider $(\cos \frac{\pi}{3}+i\sin \frac{\pi}{3})^n=\cos \frac{\pi n}{3}+i\sin \frac{\pi n}{3}=\sum_{k=0}^{n} \binom{n}{k} \Big( \frac{1}{2} \Big)^{n-k} \Big(i \frac{\sqrt{3}}{2} \Big)^{k}= \frac{1}{2^n} \sum_{k=0}^{n} \binom{n}{k} i^{k} 3^{\frac{k}{2}}$ from what we have $2^n \cos \frac{\pi n}{3} = \sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} \binom{n}{2k} (-3)^k$, but in this case $(-3)^k$ appears in the sum and I don't know how to deal with these coeffitients.
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1This is a multisection of the binomial series, apply the binomial theorem to $(1+\omega)^n$, where $\omega$ are the third roots of unity. – Maxime Jaccon Jan 04 '25 at 14:21