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Prove for all positive integers $n$ :
$\sum_{k=0}^n\left(\begin{array}{l} 3 n \\ 3 k \end{array}\right)=\frac{1}{3}\left(2^{3 n}+2 \cdot(-1)^n\right)$
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Let $\zeta_{1,2}=\frac{-1 \pm i \sqrt{3}}{2}$ be the third roots of unity, i.e., $\zeta_1^3=\zeta_2^3=1$. Then we have $\zeta_1+\zeta_2+1=0, \zeta_1^2=\zeta_2$ and $\zeta_2^2=\zeta_1$. If we plug $y=1$ and $x=1, x=\zeta_1, x=\zeta_2$ into the binomial theorem and add the equations, we obtain
$\sum_{l=0}^{3 n}\left(\begin{array}{c} 3 n \\ l \end{array}\right)\left(1+\zeta_1^l+\zeta_2^l\right)=2^{3 n}+\left(1+\zeta_1\right)^{3 n}+\left(1+\zeta_2\right)^{3 n} .$
The expression $1+\zeta_1^l+\zeta_2^l$ takes on different values depending on $l$ : for $l=3 k$, one has
$1+\zeta_1^l+\zeta_2^l=1+1^k+1^k=3,$
for $l=3 k+1$,
$1+\zeta_1^l+\zeta_2^l=1+1^k \zeta_1+1^k \zeta_2=1+\zeta_1+\zeta_2=0$

and for $l=3 k+2$,
$1+\zeta_1^l+\zeta_2^l=1+1^k \zeta_2+1^k \zeta_1=1+\zeta_2+\zeta_1=0$
hence
$\sum_{k=0}^n 3\left(\begin{array}{l} 3 n \\ 3 k \end{array}\right)=2^{3 n}+\left(1+\zeta_1\right)^{3 n}+\left(1+\zeta_2\right)^{3 n} .$
Now $1+\zeta_{1,2}=\frac{1 \pm \sqrt{3} i}{2}$ are sixth roots of unity, i.e., $\left(1+\zeta_{1,2}\right)^3=-1$ and $\left(1+\zeta_{1,2}\right)^6=1$. Hence it follows immediately that
$\sum_{k=0}^n\left(\begin{array}{l} 3 n \\ 3 k \end{array}\right)=\frac{1}{3}\left(2^{3 n}+2 \cdot(-1)^n\right) .$
by Platinum (164,234 points)

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