A twist on my last problem?

We can rewrite $C$ as

$= \displaystyle \sum_{k=0}^{1007} \omega^{2k-1}$

We can multiply the numerator and denominator by $\omega$ .

$= \dfrac{1}{\omega} \displaystyle \sum_{k=0}^{1007} \omega^{2k}$

We can see that the sum is a finite geometric progression with a common ratio $\omega^2$ and $1008$ terms.

$= \dfrac{1-(\omega^2)^{1008} }{ \omega(1-\omega^2)}$

Since $\omega = \cos \dfrac{2\pi }{2016} + i \sin \dfrac{2\pi}{2016}$ , $(\omega^2)^{1008} = \omega^{2016} = \cos 2 \pi + i \sin 2 \pi = 1$ , then the numerator of the sum is $0$ , and the sum equals $\boxed{0}$ .

In the last problem (see that problem for the proof), we established that:

$\large{1 + \omega^2 + \omega^4 + .... \omega^{2014} = 0 }$

So then a transformation gives the following:

$\large{\omega( \omega^{-1} + \omega^1 + \omega^3 + .... \omega^{2013}) = 1 + \omega^2 + \omega^4 + .... \omega^{2014} = 0 \\ \omega \neq 0 \implies \omega^{-1} + \omega^1 + \omega^3 + .... \omega^{2013} = 0}$

Note that $\omega^{2016} = 1$ . $\implies \omega^{2016-k} = \omega^{-k}$ and $\omega^{2015} = \omega^{-1}$ . Therefore,

$\begin{aligned} C & = \sum_{k=0}^{1007} \omega^{2k+1} & \small \color{#3D99F6} \text{Since } \omega^{2016-k} = \omega^{-k} \\ & = \sum_{k=0}^{503} \left(\omega^{2k+1} + \omega^{-(2k+1)} \right) & \small \color{#3D99F6} \text{and } \omega = e^{\frac \pi{1008}i} \\ & = \sum_{k=0}^{503} \left(e^{\frac {2k+1}{1008}\pi i} + e^{-\frac {2k+1}{1008}\pi i} \right) & \small \color{#3D99F6} \text{Note that } \cos \theta = \frac {e^{i\theta} + e^{-i\theta}}2 \\ & = 2 \sum_{k=0}^{503} \cos \left(\frac {2k+1}{1008}\pi \right) \\ & = 2 \left(\sum_{k=0}^{251} \cos \left(\frac {2k+1}{1008}\pi \right) + \color{#3D99F6} \sum_{k=252}^{503} \cos \left(\frac {2k+1}{1008}\pi \right) \right) & \small \color{#3D99F6} \text{and } \cos (\pi - \theta) = - \cos \theta \\ & = 2 \left(\sum_{k=0}^{251} \cos \left(\frac {2k+1}{1008}\pi \right) \color{#3D99F6} - \sum_{k=252}^{503} \cos \left(\frac {1008-(2k+1)}{1008}\pi \right) \right) \\ & = 2 \left(\sum_{k=0}^{251} \cos \left(\frac {2k+1}{1008}\pi \right) \color{#3D99F6} - \sum_{k=0}^{251} \cos \left(\frac {2k+1}{1008}\pi \right) \right) \\ & = \boxed 0 \end{aligned}$

$\omega^{2016}-1=0$

$1+\omega+\omega^2+\omega^3+... +\omega^{2015}=0 , ( \omega -1\neq 0)$

$C\omega+C\omega^2 =0$

$C(\omega+\omega^2)=0$

$C=\boxed{0} , (\omega+\omega^2 \neq 0)$