How can I calculate $\sum_{k=0}^{2n}{\binom{6n}{3k}(-1)^{k}}$

Question

how can I calculate this expression:

$\sum_{k=0}^{2n}{\binom{6n}{3k}(-1)^{k}}$ using the identity $ (x + y)^n $ = $\sum_{k=0}^{n}{\binom{n}{k}x^{n-k} y^k}$ or any other way that doesn't include Euler's number.

Answer 1

Denoting $ x=\mathrm{e}^{-\mathrm{i}\frac{\pi}{3}} $, $ y=-1 $, $ z=\mathrm{e}^{\mathrm{i}\frac{\pi}{3}} $, let $ n $ be a positive integer.

$ \left(1+x\right)^{6n}=\sum\limits_{k=0}^{6n}{\binom{6n}{k}\mathrm{e}^{-\mathrm{i}\frac{k\pi}{3}}}=\sum\limits_{k=0}^{2n}{\left(-1\right)^{k}\binom{6n}{3k}}+\mathrm{e}^{-\mathrm{i}\frac{\pi}{3}}\sum\limits_{k=0}^{2n-1}{\left(-1\right)^{k}\binom{6n}{3k+1}}+\mathrm{e}^{-\mathrm{i}\frac{2\pi}{3}}\sum\limits_{k=0}^{2n-1}{\left(-1\right)^{k}\binom{6n}{3k+2}} \cdot $

$ \left(1+y\right)^{6n}=\sum\limits_{k=0}^{6n}{\left(-1\right)^{k}\binom{6n}{k}}=\sum\limits_{k=0}^{2n}{\left(-1\right)^{k}\binom{6n}{3k}}-\sum\limits_{k=0}^{2n-1}{\left(-1\right)^{k}\binom{6n}{3k+1}}+\sum\limits_{k=0}^{2n-1}{\left(-1\right)^{k}\binom{6n}{3k+2}} \cdot $

$ \left(1+z\right)^{n}=\sum\limits_{k=0}^{6n}{\binom{6n}{k}\mathrm{e}^{\mathrm{i}\frac{k\pi}{3}}}=\sum\limits_{k=0}^{2n}{\left(-1\right)^{k}\binom{6n}{3k}}+\mathrm{e}^{\mathrm{i}\frac{\pi}{3}}\sum\limits_{k=0}^{2n-1}{\left(-1\right)^{k}\binom{6n}{3k+1}}+\mathrm{e}^{\mathrm{i}\frac{2\pi}{3}}\sum\limits_{k=0}^{2n-1}{\left(-1\right)^{k}\binom{6n}{3k+2}} $

Using the fact that $ x+y+z=0 $ and that $ x^{2}+y^{2}+z^{2}=0 $, summing the previous relations, we get : $$ \left(1+x\right)^{6x}+\left(1+y\right)^{6n}+\left(1+z\right)^{6n}=3\sum_{k=0}^{2n}{\left(-1\right)^{k}\binom{6n}{3k}} $$

Since \begin{aligned} \left(1+x\right)^{6x}+\left(1+y\right)^{6n}+\left(1+z\right)^{6n}&=\left(\mathrm{e}^{-\mathrm{i}\frac{\pi}{6}}\left(\mathrm{e}^{\mathrm{i}\frac{\pi}{6}}+\mathrm{e}^{-\mathrm{i}\frac{\pi}{6}}\right)\right)^{6n}+\left(\mathrm{e}^{\mathrm{i}\frac{\pi}{6}}\left(\mathrm{e}^{\mathrm{i}\frac{\pi}{6}}+\mathrm{e}^{-\mathrm{i}\frac{\pi}{6}}\right)\right)^{6n}\\ &=2^{6n+1}\left(-1\right)^{n}\cos^{6n}{\left(\frac{\pi}{6}\right)}\\ &=2\left(-1\right)^{n}3^{3n} \end{aligned}

We get $$ \sum_{k=0}^{2n}{\left(-1\right)^{k}\binom{6n}{3k}}=2\left(-1\right)^{n}3^{3n-1} $$