p-0-n-psh-a-bp-

Question Number 147828 by puissant last updated on 23/Jul/21

\underset{p = 0}{\sum^{n}} psh (a + bp)

Answered by Olaf_Thorendsen last updated on 23/Jul/21

\underset{p = 0}{\sum^{n}} ch (2 a + 2 p b) = \frac{ch (2 a + n b) sh ((n + 1) b)}{sh b}

(voir question precedente)

\Rightarrow \underset{p = 0}{\sum^{n}} ch (a + p b) = \frac{ch (a + \frac{n}{2} b) sh (\frac{n + 1}{2} b)}{sh \frac{b}{2}}

\frac{\partial}{\partial b} \underset{p = 0}{\sum^{n}} ch (a + p b) = \underset{p = 0}{\sum^{n}} p sh (a + p b)

= \frac{\partial}{\partial b} (\frac{ch (a + \frac{n}{2} b) sh (\frac{n + 1}{2} b)}{sh \frac{b}{2}})

\dots calcul fastidieux!

Commented by puissant last updated on 23/Jul/21

merci prof c^{'} est trivial \dots