<s>[INST] Divisibility of Sum of Equally Spaced Binomial Coefficients According to a numerical calculation I did for small values of $k$, it appears that the following is true.
$$4|\left[\sum_{j=1}^{n-1}\binom{3n}{3j}\right]$$  or $$\sum_{j=1}^{n-1}\binom{3n}{3j}=4p, p\in\mathbb{Z}$$
Ex. 
If $n=2, \binom{6}{3}=20=4\cdot 5$
If $n=3, \binom{9}{3}+\binom{9}{6}=2\cdot 84=168=4\cdot 42$
If $n=4, \binom{12}{3}+\binom{12}{6}+\binom{12}{9}=2\cdot 220+924=1364=4\cdot341$
If $n=5, \binom{15}{3}+\binom{15}{6}+\binom{15}{9}+\binom{15}{12}=2\cdot 455+2\cdot 5005=10920=4\cdot 2730$
Is there a way to prove this?  Using induction, as above I've shown the base case is true.  Then if we assume that 
$$S_m=\sum_{j=1}^{m-1}\binom{3m}{3j}=4q, q\in\mathbb{Z}$$
Then 
$$S_{m+1}=\sum_{j=1}^{m}\binom{3m+3}{3j}=?$$
And I have no idea how to go forward.  Perhaps its not true?  Is there a counterexample?
 [/INST] Starting with the binomial theorem,
$$(1+x)^{3m} = \sum_{k=0}^{3m} \binom{3m}{k} x^k,$$
we see that for the cube root of unity $\omega = e^{2\pi i/3}$,
$$(1+x)^{3m} + (1+\omega x)^{3m} + (1+\omega^2 x)^{3m} = 3 \sum_{\substack{k=0 \\ k \equiv 0 \pmod{3}}}^{3m} \binom{3m}{k} x^k.$$
The terms where $k \not\equiv 0 \pmod{3}$ are sifted away, since $1 + \omega + \omega^2 = 0$.  Now evaluate at $x=1$:
$$2^{3m} + (1+\omega)^{3m} + (1+\omega^2)^{3m} = 3 \sum_{\substack{k=0 \\ k \equiv 0 \pmod{3}}}^{3m} \binom{3m}{k}.$$
Since $1+\omega = e^{\pi i/3}$ and $1+\omega^2 = e^{-\pi i/3}$ are 6-th roots of unity, it follows that $(1+\omega)^3 = (1+\omega^2)^3 = -1$.
The right-hand side is almost 3 times your sum $S_m$, but $S_m$ does not include the $k=0$ and $k=3m$ terms.  Subtracting them away, we find
$$S_m = \frac{2^{3m} + 2((-1)^m-3)}{3}.$$
For $m \geqslant 1$, the numerator is always divisible by 4, which explains the observed divisibility.
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