 Solution

First establish the binomial theorem needed throughout this question. For integers $0\le m\le n$, define the binomial coefficient $\left(\genfrac{}{}{0ex}{}{n}{m}\right)$ to be the number of $m$-element subsets of an $n$-element set. Counting ordered selections and then forgetting their order giveswith the factorial convention $0!=1$. In the product of $n$ factors $(1+z)$, choose $z$ from exactly $m$ factors and $1$ from the others. Each choice contributes $z^m$, and there are $\left(\genfrac{}{}{0ex}{}{n}{m}\right)$ choices. The distributive law therefore provesfor every complex number $z$, including $n=0$ by the empty-product convention.

To evaluate the even-indexed sum, use the even part of a polynomial: averaging the binomial theorem at $z$ and $-z$ cancels the odd powers. For $z=i\sqrt{3}$,The two complex numbers satisfy $1\pm i\sqrt{3}=2e^{\pm i\pi /3}$. Their $6n$th powers are both $2^{6n}$, since $e^{\pm 2\pi in}=1$. Thus the sum is