The sum $\sum_{k=0}^{3n}(-3)^k \binom{6n}{2k}=2^{6n}$ The question asks to show that $$\sum_{k=0}^{3n}(-3)^k \binom{6n}{2k}=2^{6n}$$ by considering the binomial expansion 
I thought about the use of $$(1+z)^n=\sum_{k=0}^{n}\binom{n}{k}z^k$$
with suitable complex number $z$, as the formula shows the $(-3)^k$ term might suggest the use of complex number, which take the imaginary part of the expansion
However, I cannot find such $z$ that makes the sum to $2^{6n}$
Any hints are appreciated!
 A: Consider the real part of the binomial expansion of $(1+i\sqrt{3})^{6n}=(2e^{i\pi/3})^{6n}=2^{6n}$:
$$2^{6n}=\mbox{Re}\left((1+i\sqrt{3})^{6n}\right)=
\mbox{Re}\left(\sum_{j = 0}^{6n}{6n \choose j}(i\sqrt{3})^{j} \right)
=\sum_{k = 0}^{3n}{6n \choose 2k}(-3)^k.$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
\sum_{k = 0}^{3n}\pars{-3}^{k}{6n \choose 2k} & =
\sum_{k = 0}^{\infty}{6n \choose 2k}\pars{\root{3}\ic}^{2k} =
\sum_{k = 0}^{\infty}{6n \choose k}\pars{\root{3}\ic}^{k}
\,\,{1 + \pars{-1}^{k} \over 2} \\[5mm] & =
{1 \over 2}\sum_{k = 0}^{\infty}{6n \choose k}\pars{\root{3}\ic}^{k} +
{1 \over 2}\sum_{k = 0}^{\infty}{6n \choose k}\pars{-\root{3}\ic}^{k}
\\[5mm] & =
\Re\pars{\sum_{k = 0}^{\infty}{6n \choose k}\bracks{\root{3}\ic}^{k}}
=
\Re\pars{\bracks{1 + \root{3}\ic}^{6n}}= \pars{1 + 3}^{3n}
\\[5mm] & =
\bbx{\ds{2^{6n}}}
\end{align}
