Prove that \displaystyle{\underset{{{r}={0}}}{{\overset{{{n}}}{\sum}}}}.^{{{n}}}{C}_{{{r}}}.{\left({n}-{r}\right)}{\cos{{\left(\frac{{{2}{r}\pi}}{{{n}}}\right)}}}=-{n}{.2}^{{{n}-{1}}}.{{\cos}^{{{n}}}'}\frac{{\pi}}{{{n}}}.

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Prove that \displaystyle{\underset{{{r}={0}}}{{\overset{{{n}}}{\sum}}}}.^{{{n}}}{C}_{{{r}}}.{\left({n}-{r}\right)}{\cos{{\left(\frac{{{2}{r}\pi}}{{{n}}}\right)}}}=-{n}{.2}^{{{n}-{1}}}.{{\cos}^{{{n}}}'}\frac{{\pi}}{{{n}}}.

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Question Text

Prove that \displaystyle{\underset{{{r}={0}}}{{\overset{{{n}}}{\sum}}}}.^{{{n}}}{C}_{{{r}}}.{\left({n}-{r}\right)}{\cos{{\left(\frac{{{2}{r}\pi}}{{{n}}}\right)}}}=-{n}{.2}^{{{n}-{1}}}.{{\cos}^{{{n}}}'}\frac{{\pi}}{{{n}}}.

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