- k - 1 6 - sin 2 k - 7 - i cos 2 k - 7 - -

The correct option is B i i ∑ _ k = 1 ^ 6 ( cos 2 k π/ 7 + i sin 2 k π/ 7 ) i [ 1 + α + α ^ 2 + … + α ^ 6 1 ] = i [ 0 1 ] = i ...

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∑ k = 1 6 [...

Question

$\sum_{k = 1}^{6} [sin \frac{2 k π}{7} - i cos \frac{2 k π}{7}] =$

- 1

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6 \sum k = 1 [sin \frac{2 k π}{7} - i cos \frac{2 k π}{7}] =

6 \sum k = 1 [sin \frac{2 k π}{7} - i cos \frac{2 k π}{7}]

6 \sum k = 1 [sin \frac{2 k π}{7} - i cos \frac{2 k π}{7}]

7 \sum k = 1 [sin \frac{2 k Π}{7} - i cos \frac{2 k Π}{7}]

\sum_{k = 1}^{4} [sin \frac{2 k π}{5} - i cos \frac{2 k π}{5}]

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