How do I prove ${\binom{n}{0}}+{\binom{n}{3}}+{\binom{n}{6}}+\cdots ={\frac{1}{3}}\left(2^{n}+2\cos {\frac{n\pi }{3}}\right)?$

Question

I think it is solved with double counting. I tried to write unfolded, but I failed to solve the exercise. Can you help me?

Hint: Let $\omega = e^{i\frac{2\pi}{3}}$, we have $1 + \omega^k + \omega^{2k} = \begin{cases}3, & k \equiv 0 \pmod 3\0, &k \not\equiv 0\pmod 3\end{cases}$, now apply binomial theorem to $(1+1)^n + (1+\omega)^n + (1+\omega^2)^n$, what do you get? — achille hui, Oct 14 '20 at 20:31

score 1 · Answer 1 · answered Oct 14 '20 at 20:27

1

that is a multisection of the binomial series

answered Oct 14 '20 at 20:27

G Cab

1 Answers1