An algebra problem by Hobart Pao

By GP formula:

$\large \displaystyle C = \frac{w^{2014} w^2 - 1}{w^2-1}$

$\large \displaystyle C = \frac{w^{2016} - 1}{w^2-1}$

$\large \displaystyle C = \frac{1 - 1}{w^2-1}$

$\color{#3D99F6}\boxed{ \large \displaystyle C = 0}$

We can rewrite $C$ as

$= \displaystyle \sum_{k=0}^{1007} (\omega^2)^k$

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

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

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 $\boxed{0}$ .

Multiplying C by ω², we get C*ω² = C using the fact that ω is a 2016th root of unity. Since ω is thus also not 0, we must have C = 0.