The sum $\sum_{k=0}^{3n}(-3)^k \binom{6n}{2k}=2^{6n}$

Question

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!

Answer 1

Consider the real part of the binomial expansion of $(1+i\sqrt{3})^{6n}=(2e^{i\pi/3})^{6n}=2^{6n}$: $$2^{6n}=\mbox{Re}\left((1+i\sqrt{3})^{6n}\right)= \mbox{Re}\left(\sum_{j = 0}^{6n}{6n \choose j}(i\sqrt{3})^{j} \right) =\sum_{k = 0}^{3n}{6n \choose 2k}(-3)^k.$$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}