∑ nCr * (n-r) * cos{(2*r*pi)/n} where ‘r’ ranges from ‘0’ to ‘n’ = ??

Question

Sourabh Singh · Answer

Hiii Use cos in terms of Exponential adn then separate out the terms. Then operate individually you will get the answer.

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∑ nCr * (n-r) * cos{(2rpi)/n} where ‘r’ ranges from ‘0’ to ‘n’ = ??

∑ nCr * (n-r) * cos{(2rpi)/n}
where ‘r’ ranges from ‘0’ to ‘n’ = ??

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∑ nCr * (n-r) * cos{(2*r*pi)/n} where ‘r’ ranges from ‘0’ to ‘n’ = ??

∑ nCr * (n-r) * cos{(2*r*pi)/n} where ‘r’ ranges from ‘0’ to ‘n’ = ??

1 Answers

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∑ nCr * (n-r) * cos{(2rpi)/n} where ‘r’ ranges from ‘0’ to ‘n’ = ??

∑ nCr * (n-r) * cos{(2rpi)/n}
where ‘r’ ranges from ‘0’ to ‘n’ = ??