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

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

Answered by Olaf_Thorendsen last updated on 23/Jul/21

$$$$$$\underset{p=0}{\sum ^n}\mathrm{ch}(2a+2pb)=\frac{\mathrm{ch}(2a+nb)\mathrm{sh}\left((n+1)b\right)}{\mathrm{sh}b}$$$$\left(\mathrm{voir}\mathrm{question}\mathrm{precedente}\right)$$$$\Rightarrow \underset{p=0}{\sum ^n}\mathrm{ch}(a+pb)=\frac{\mathrm{ch}(a+\frac{n}{2}b)\mathrm{sh}\left(\frac{n+1}{2}b\right)}{\mathrm{sh}\frac{b}{2}}$$$$\frac{\partial }{\partial b}\underset{p=0}{\sum ^n}\mathrm{ch}(a+pb)=\underset{p=0}{\sum ^n}p\mathrm{sh}(a+pb)$$$$$$$$=\frac{\partial }{\partial b}\left(\frac{\mathrm{ch}(a+\frac{n}{2}b)\mathrm{sh}\left(\frac{n+1}{2}b\right)}{\mathrm{sh}\frac{b}{2}}\right)$$$$...\mathrm{calcul}\mathrm{fastidieux}!$$

Commented by puissant last updated on 23/Jul/21

$$\mathrm{merci}\mathrm{prof}{\mathrm{c}}^{\prime }\mathrm{est}\mathrm{trivial}...$$

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