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Question Number 167421 by infinityaction last updated on 16/Mar/22

	Answered by mindispower last updated on 16/Mar/22

	${c o t}^{2} (\frac{k π}{2 n + 1}) = {c o t}^{2} (s π) = {(i \frac{e^{i s} + e^{- i s}}{e^{i s} - e^{- i s}})}^{2} =$ $Missing \left or extra \right$ $\Rightarrow i c o t (s π) p = \frac{e^{2 i s} + 1}{e^{2 i s} - 1}$ $\Rightarrow e^{2 i s} = \frac{1 + i a}{i a - 1}, ({(e^{2 i s})}^{2 n + 1} - 1) = 0$ $\Rightarrow {(1 + i a)}^{2 n + 1} + {(1 - i a)}^{2 n + 1} = 0$ $\Rightarrow \underset{k = 0}{\sum^{2 n + 1}} (\begin{matrix} 2 n + 1 \\ k \end{matrix}) ({(i a)}^{k} + {(- i a)}^{k}) = 0$ $\Rightarrow \underset{k = 0}{\sum^{n}} (\begin{matrix} 2 n + 1 \\ 2 k \end{matrix}) . 2 {(- 1)}^{k} a^{2 k} = 0$ $a^{2} = {t g}^{2} (\frac{s π}{2 n + 1}), f o r s o m s \in [1, n]$ $\Rightarrow {t g}^{2} (\frac{s π}{2 n + 1}) R o o t O f$ $2 \underset{k = 0}{\sum^{n}} (\begin{matrix} 2 n + 1 \\ 2 k \end{matrix}) {(- 1)}^{k} X^{k} = 0$ ${(- 1)}^{n} (\begin{matrix} 2 n + 1 \\ 1 \end{matrix}) X^{n} + \underset{k = 0}{\sum^{n - 1}} {(- 1)}^{k} (\begin{matrix} 2 n + 1 \\ k \end{matrix}) X^{k} = 0$ $\Leftrightarrow X^{n} + \underset{k = 0}{\sum^{n - 1}} \frac{{(- 1)}^{k - n} (\begin{matrix} 2 n + 1 \\ k \end{matrix})}{(\begin{matrix} 2 n + 1 \\ 1 \end{matrix})} X^{k} = 0$
	Commented by infinityaction last updated on 16/Mar/22

	$t h a n k y o u s i r$
	Commented by infinityaction last updated on 16/Mar/22

	$s i r t h e s e c o n d p a r t o f t h i s a l s o h a s t o b e s o l v e d$
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