Sum of series $1−ω^2+ω^4−ω^6+ω^8−ω^{10}+ω^{12}+⋯+ω^{600}−ω^{602}+ω^{604}$

Question

I need to find sum of the series involving cube roots of unity $1−ω^2+ω^4−ω^6+ω^8−ω^{10}+ω^{12}+⋯+ω^{600}−ω^{602}+ω^{604}$.

Found it in an old test paper.

I applied Geometric Progression Sum Formula.

$$\frac{1.((-ω^2)^{(303+1)}-1)}{-ω^2-1}$$

$$=\frac{(ω^{608}-1)}{-ω^2-1}$$

$$=\frac{(ω^{2}-1)}{-ω^2-1}$$

$$=\frac{(ω^{2}-1)}{ω}$$

$$=(ω-ω^2)$$

$$=(-1-2ω^2)$$

But the answer given is $-2ω^2$ only.Don't know how.Can anyone spot any mistake in what I did?

P.S:Thanks guys.I found my mistake now.Applied the formula wrong in the first step.It should be

$$\frac{1.((-ω^2)^{(302+1)}-1)}{-ω^2-1}$$

Answer 1

$$\sum_{r=0}^{302}(-\omega^2)^r=1\cdot\dfrac{1-(-\omega^2)^{303}}{1-(-\omega^2)}=\dfrac{1+(\omega^3)^{202}}{1+\omega^2}=? $$