<s>[INST] Finding $\binom{n}{0} + \binom{n}{3} + \binom{n}{6} + \ldots $ Help me to simplify:$$\binom{n}{0} + \binom{n}{3} + \binom{n}{6} + \ldots $$
I got a hunch that it will depend on whether $n$ is a multiple of $6$ and equals to $\frac{2^n+2}{3}$ when $n$ is a multiple of $6$.
 [/INST] If $1+\zeta+\zeta^2=0$, try to compute $(1+\zeta^0)^n + (1+\zeta)^n + (1+\zeta^2)^n$. Hint: $\zeta^3=1$, and, if $k$ is not divisible by $3$, $1+\zeta^k+\zeta^{2k}=0$.
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