Question Number 147819 by puissant last updated on 23/Jul/21
$$\underset{\mathrm{p}=\mathrm{0}} {\overset{\mathrm{n}} {\sum}}\mathrm{ch}^{\mathrm{2}} \left(\mathrm{a}+\mathrm{pb}\right) \\ $$
Answered by Olaf_Thorendsen last updated on 23/Jul/21
$$\mathrm{S}\:=\:\underset{{p}=\mathrm{0}} {\overset{{n}} {\sum}}\mathrm{ch}^{\mathrm{2}} \left(\mathrm{a}+\mathrm{p}{b}\right) \\ $$$$\mathrm{S}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\underset{{p}=\mathrm{0}} {\overset{{n}} {\sum}}\left(\mathrm{1}+\mathrm{ch}\left(\mathrm{2a}+\mathrm{2p}{b}\right)\right) \\ $$$$\mathrm{S}\:=\:\frac{{n}+\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\underset{{p}=\mathrm{0}} {\overset{{n}} {\sum}}\mathrm{ch}\left(\mathrm{2}{a}+\mathrm{2p}{b}\right)\:\:\:\left(\mathrm{1}\right) \\ $$$$ \\ $$$$\:\mathrm{ch}\left(\mathrm{2}{a}+\mathrm{2}{pb}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left({e}^{\mathrm{2}{a}} {e}^{\mathrm{2}{pb}} +{e}^{−\mathrm{2}{a}} {e}^{−\mathrm{2}{pb}} \right) \\ $$$$\:\underset{{p}=\mathrm{0}} {\overset{{n}} {\sum}}\mathrm{ch}\left(\mathrm{2}{a}+\mathrm{2}{pb}\right) \\ $$$$=\:\frac{{e}^{\mathrm{2}{a}} }{\mathrm{2}}\underset{{p}=\mathrm{0}} {\overset{{n}} {\sum}}{e}^{\mathrm{2}{pb}} +\frac{{e}^{−\mathrm{2}{a}} }{\mathrm{2}}\underset{{p}=\mathrm{0}} {\overset{{n}} {\sum}}{e}^{−\mathrm{2}{pb}} \\ $$$$\:=\:\:\frac{{e}^{\mathrm{2}{a}} }{\mathrm{2}}.\frac{\mathrm{1}−{e}^{\mathrm{2}\left({n}+\mathrm{1}\right){b}} }{\mathrm{1}−{e}^{\mathrm{2}{b}} }+\frac{{e}^{−\mathrm{2}{a}} }{\mathrm{2}}.\frac{\mathrm{1}−{e}^{−\mathrm{2}\left({n}+\mathrm{1}\right){b}} }{\mathrm{1}−{e}^{−\mathrm{2}{b}} } \\ $$$$\:=\:\:\frac{{e}^{\mathrm{2}{a}} {e}^{{nb}} }{\mathrm{2}}.\frac{{e}^{−\left({n}+\mathrm{1}\right){b}} −{e}^{\left({n}+\mathrm{1}\right){b}} }{{e}^{−{b}} −{e}^{{b}} }+\frac{{e}^{−\mathrm{2}{a}} {e}^{−{nb}} }{\mathrm{2}}.\frac{{e}^{\left({n}+\mathrm{1}\right){b}} −{e}^{−\left({n}+\mathrm{1}\right){b}} }{{e}^{{b}} −{e}^{−{b}} } \\ $$$$\:=\:\:\frac{{e}^{\mathrm{2}{a}+{nb}} }{\mathrm{2}}.\frac{\mathrm{sh}\left(\left({n}+\mathrm{1}\right){b}\right)}{\mathrm{sh}{b}}+\frac{{e}^{−\left(\mathrm{2}{a}+{nb}\right)} }{\mathrm{2}}.\frac{\mathrm{sh}\left(\left({n}+\mathrm{1}\right){b}\right)}{\mathrm{sh}{b}} \\ $$$$\:=\:\:\frac{\mathrm{ch}\left(\mathrm{2}{a}+{nb}\right)\mathrm{sh}\left(\left({n}+\mathrm{1}\right){b}\right)}{\mathrm{sh}{b}} \\ $$$$ \\ $$$$\left(\mathrm{1}\right)\::\:\mathrm{S}\:=\:\frac{{n}+\mathrm{1}}{\mathrm{2}}+\:\frac{\mathrm{ch}\left(\mathrm{2}{a}+{nb}\right)\mathrm{sh}\left(\left({n}+\mathrm{1}\right){b}\right)}{\mathrm{2sh}{b}} \\ $$
Commented by puissant last updated on 23/Jul/21
$$\mathrm{jolie}\:\mathrm{prof}\:\mathrm{merci}.. \\ $$