Complex Binomial !

Relevant wiki: Euler's Formula

Using binomial theorem ,

$\begin{aligned} \sum_{k=0}^{2n} \binom {2n}k x^k & = (1+x)^{2n} \\ \sum_{k=0}^{2n} \binom {2n}k (-x)^k & = (1-x)^{2n} \\ \sum_{k=0}^{2n} \binom {2n}k x^k - \sum_{k=0}^{2n} \binom {2n}k (-x)^k & = (1+x)^{2n} - (1-x)^{2n} \\ 2 \sum_{k=1}^n \binom {2n}{2k-1} x^{2k-1} & = (1+x)^{2n} - (1-x)^{2n} \\ \frac 2x \sum_{k=1}^n \binom {2n}{2k-1} (x^2)^k & = (1+x)^{2n} - (1-x)^{2n} \\ \sum_{k=1}^n \binom {2n}{2k-1} (x^2)^k & = \frac x2 \left((1+x)^{2n} - (1-x)^{2n}\right) & \small \color{#3D99F6} \text{Putting }n = 999, x = \sqrt 3i \\ \sum_{k=1}^{999} \binom {1998}{2k-1} (-3)^k & = \frac {\sqrt 3}2i \color{#3D99F6} \left((1+\sqrt 3i)^{1998} - (1-\sqrt 3 i)^{1998}\right) & \small \color{#3D99F6} \text{By Euler's formula} \\ & = \frac {\sqrt 3}2 i \color{#3D99F6} \left(2^{1998}\left(e^{1998\cdot \frac \pi 3 i} - e^{-1998\cdot \frac \pi 3 i}\right) \right) \\ & = - \sqrt 3\cdot 2^{1998} \sin 666 \pi \\ & = \boxed{0} \end{aligned}$

$\displaystyle\sum_{k=1}^{999} \binom{1998}{2k-1} (-3)^k =$ Imaginary part of $\sqrt{3}(1-\sqrt{3}i)^{1998}=0$ .