Σ_(p=0) ^n ch^2 (a+pb)


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Question Number 147819 by puissant last updated on 23/Jul/21

$\underset{p = 0}{\sum^{n}} {ch}^{2} (a + pb)$
	Answered by Olaf_Thorendsen last updated on 23/Jul/21

	$S = \underset{p = 0}{\sum^{n}} {ch}^{2} (a + p b)$ $S = \frac{1}{2} \underset{p = 0}{\sum^{n}} (1 + ch (2 a + 2 p b))$ $S = \frac{n + 1}{2} + \frac{1}{2} \underset{p = 0}{\sum^{n}} ch (2 a + 2 p b) (1)$ $ch (2 a + 2 p b) = \frac{1}{2} (e^{2 a} e^{2 p b} + e^{- 2 a} e^{- 2 p b})$ $\underset{p = 0}{\sum^{n}} ch (2 a + 2 p b)$ $= \frac{e^{2 a}}{2} \underset{p = 0}{\sum^{n}} e^{2 p b} + \frac{e^{- 2 a}}{2} \underset{p = 0}{\sum^{n}} e^{- 2 p b}$ $= \frac{e^{2 a}}{2} . \frac{1 - e^{2 (n + 1) b}}{1 - e^{2 b}} + \frac{e^{- 2 a}}{2} . \frac{1 - e^{- 2 (n + 1) b}}{1 - e^{- 2 b}}$ $= \frac{e^{2 a} e^{n b}}{2} . \frac{e^{- (n + 1) b} - e^{(n + 1) b}}{e^{- b} - e^{b}} + \frac{e^{- 2 a} e^{- n b}}{2} . \frac{e^{(n + 1) b} - e^{- (n + 1) b}}{e^{b} - e^{- b}}$ $= \frac{e^{2 a + n b}}{2} . \frac{sh ((n + 1) b)}{sh b} + \frac{e^{- (2 a + n b)}}{2} . \frac{sh ((n + 1) b)}{sh b}$ $= \frac{ch (2 a + n b) sh ((n + 1) b)}{sh b}$ $(1) : S = \frac{n + 1}{2} + \frac{ch (2 a + n b) sh ((n + 1) b)}{2 sh b}$
	Commented by puissant last updated on 23/Jul/21

	$jolie prof merci . .$
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