k-0-n-1-sec-2-kpi-n-n-2-

Question Number 142393 by qaz last updated on 31/May/21

\underset{k = 0}{\sum^{n - 1}} \sec^{2} (\frac{k π}{n}) = n^{2} \dots \dots ? ? ?

Answered by mindispower last updated on 31/May/21

= \underset{k = 0}{\sum^{n - 1}} (1 + {t g}^{2} (\frac{k π}{n})) = n + S

S = \underset{k = 0}{\sum^{n - 1}} {t g}^{2} (\frac{k π}{n}) = \underset{k = 1}{\sum^{n - 1}} \frac{1}{{c o t}^{2} (\frac{k π}{n})} w i t h e k \neq \frac{n}{2}, n \notin 2 N

n = 2 m + 1

t g (x) = \frac{e^{i x} - e^{- i x}}{i (e^{i x} + e^{- i x})}

l e t {(Z - 1)}^{n} - {(Z + 1)}^{n} = P (Z) = 0

\Rightarrow \frac{Z - 1}{Z + 1} = e^{2 i k \frac{π}{n}}

Z = \frac{e^{\frac{2 i k π}{n}} + 1}{1 - e^{\frac{2 i k π}{n}}} = i c o t (\frac{k π}{n}), k \in [1, n - 1]

W e h a v e \frac{P^{'}}{P} = \underset{k = 1}{\sum^{n - 1}} \frac{1}{X - i c o t (\frac{k π}{n})}

\Rightarrow \frac{{P P}^{″} - p {(x)}^{' 2}}{p^{2} (x)} = \underset{k = 1}{\sum^{n - 1}} - \frac{1}{{(x - i c o t (\frac{k π}{n}))}^{2}}

p (0) = - 2

p^{'} (0) = 0, n = 2 m + 1

p^{″} (z) = n (n - 1) {(z - 1)}^{n - 2} - n (n - 1) {(z + 1)}^{n - 2}

= - 2 n (n - 1)

\frac{4 n (n - 1)}{4} = \underset{k = 1}{\sum^{n - 1}} \frac{1}{{c o t}^{2} (\frac{k π}{n})} = n^{2} - n

\underset{k = 0}{\sum^{n - 1}} {s e c}^{2} (\frac{k π}{n}) = n + S = n + n^{2} - n = n^{2}

\Leftrightarrow \underset{k = 0}{\sum^{n - 1}} {s e c}^{2} (\frac{k π}{n}) = n^{2}

Commented by qaz last updated on 31/May/21

thank you Sir power

Commented by mindispower last updated on 01/Jun/21

p l e a s u r