If omega is a non-real cube root of unity,the | vedclass

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(A) Let $S = \sum_{n=1}^{2017} (n-\omega)(n-\omega^2)$.Since $\omega^2+\omega+1=0$,we have $(n-\omega)(n-\omega^2) = n^2 - n(\omega+\omega^2) + \omega

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$-1$

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(A) Let $S = \sum_{n=1}^{2017} (n-\omega)(n-\omega^2)$.Since $\omega^2+\omega+1=0$,we have $(n-\omega)(n-\omega^2) = n^2 - n(\omega+\omega^2) + \omega^3 = n^2 + n + 1$.Thus,$S = \sum_{n=1}^{2017} (n^2 + n + 1) = \sum_{n=1}^{2017} n^2 + \sum_{n=1}^{2017} n + \sum_{n=1}^{2017} 1$.Using standard summation formulas: $S = \frac{2017(2018)(4035)}{6} + \frac{2017(2018)}{2} + 2017$.Dividing by $2017$: $\frac{S}{2017} = \frac{2018 \cdot 4035}{6} + \frac{2018}{2} + 1 = 1009 \cdot 1345 + 1009 + 1 = 1009(1345+1) + 1 = 1009(1346) + 1$.Since $1009(1346)$ is even,$\frac{S}{2017} = 	ext{even} + 1 = 	ext{odd}$.Therefore,$\cos\left( \frac{S\pi}{2017} ight) = \cos(	ext{odd} \cdot \pi) = -1$.

If $\omega$ is a non-real cube root of unity,then the value of $\cos \left[ \left\{ (1-\omega)(1-\omega^2) + (2-\omega)(2-\omega^2) + \dots + (2017-\omega)(2017-\omega^2) \right\} \cdot \frac{\pi}{2017} \right]$ is -

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