How should one prove that $\binom n{0} + \binom n{3} + \binom n{6} +... +\binom n{3k} = \dfrac{2^n-2}{3}$ for $n$ odd and $\binom n{0} + \binom n{3} + \binom n{6} +... +\binom n{3k} = \dfrac{2^n+2}{3}$ for $n$ even, where $k=\lfloor \dfrac{n}{3}\rfloor $?
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2Possible duplicate of Show that $\sum_{k=0}^n\binom{3n}{3k}=\frac{8^n+2(-1)^n}{3}$ – Tacet Nov 02 '15 at 21:37
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https://math.stackexchange.com/questions/3384818/how-to-calculate-the-binomial-sum-s-mathop-sum-limits-i-0-lfloor-fracn/3384901#3384901 – lab bhattacharjee Oct 08 '19 at 07:45
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Hint: consider $$ (1+x)^n = \sum_{k = 0}^n \binom nk x^k $$ for $x\in \{1,j,j^2\} \ \ \ (j^3 = 1, j\notin \Bbb R)$ and $1 + j + j^2 = 0$.
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