p-0-n-ch-2-a-pb-

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

S = Σ_(p=0) ^n ch^2 (a+pb) S = (1/2)Σ_(p=0) ^n (1+ch(2a+2pb)) S = ((n+1)/2)+(1/2)Σ_(p=0) ^n ch(2a+2pb) (1) ch(2a+2pb) = (1/2)(e^(2a) e^(2pb) +e^(−2a) e^(−2pb) ) Σ_(p=0) ^n ch(2a+2pb) = (e^(2a) /2)Σ_(p=0) ^n e^(2pb) +(e^(−2a) /2)Σ_(p=0) ^n e^(−2pb) = (e^(2a) /2).((1−e^(2(n+1)b) )/(1−e^(2b) ))+(e^(−2a) /2).((1−e^(−2(n+1)b) )/(1−e^(−2b) )) = ((e^(2a) e^(nb) )/2).((e^(−(n+1)b) −e^((n+1)b) )/(e^(−b) −e^b ))+((e^(−2a) e^(−nb) )/2).((e^((n+1)b) −e^(−(n+1)b) )/(e^b −e^(−b) )) = (e^(2a+nb) /2).((sh((n+1)b))/(shb))+(e^(−(2a+nb)) /2).((sh((n+1)b))/(shb)) = ((ch(2a+nb)sh((n+1)b))/(shb)) (1) : S = ((n+1)/2)+ ((ch(2a+nb)sh((n+1)b))/(2shb))

$$\mathrm{S}\:=\:\underset{{p}=\mathrm{0}} {\overset{{n}} {\sum}}\mathrm{ch}^{\mathrm{2}} \left(\mathrm{a}+\mathrm{p}{b}\right) \\ $$$$\mathrm{S}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\underset{{p}=\mathrm{0}} {\overset{{n}} {\sum}}\left(\mathrm{1}+\mathrm{ch}\left(\mathrm{2a}+\mathrm{2p}{b}\right)\right) \\ $$$$\mathrm{S}\:=\:\frac{{n}+\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\underset{{p}=\mathrm{0}} {\overset{{n}} {\sum}}\mathrm{ch}\left(\mathrm{2}{a}+\mathrm{2p}{b}\right)\:\:\:\left(\mathrm{1}\right) \\ $$$$ \\ $$$$\:\mathrm{ch}\left(\mathrm{2}{a}+\mathrm{2}{pb}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left({e}^{\mathrm{2}{a}} {e}^{\mathrm{2}{pb}} +{e}^{−\mathrm{2}{a}} {e}^{−\mathrm{2}{pb}} \right) \\ $$$$\:\underset{{p}=\mathrm{0}} {\overset{{n}} {\sum}}\mathrm{ch}\left(\mathrm{2}{a}+\mathrm{2}{pb}\right) \\ $$$$=\:\frac{{e}^{\mathrm{2}{a}} }{\mathrm{2}}\underset{{p}=\mathrm{0}} {\overset{{n}} {\sum}}{e}^{\mathrm{2}{pb}} +\frac{{e}^{−\mathrm{2}{a}} }{\mathrm{2}}\underset{{p}=\mathrm{0}} {\overset{{n}} {\sum}}{e}^{−\mathrm{2}{pb}} \\ $$$$\:=\:\:\frac{{e}^{\mathrm{2}{a}} }{\mathrm{2}}.\frac{\mathrm{1}−{e}^{\mathrm{2}\left({n}+\mathrm{1}\right){b}} }{\mathrm{1}−{e}^{\mathrm{2}{b}} }+\frac{{e}^{−\mathrm{2}{a}} }{\mathrm{2}}.\frac{\mathrm{1}−{e}^{−\mathrm{2}\left({n}+\mathrm{1}\right){b}} }{\mathrm{1}−{e}^{−\mathrm{2}{b}} } \\ $$$$\:=\:\:\frac{{e}^{\mathrm{2}{a}} {e}^{{nb}} }{\mathrm{2}}.\frac{{e}^{−\left({n}+\mathrm{1}\right){b}} −{e}^{\left({n}+\mathrm{1}\right){b}} }{{e}^{−{b}} −{e}^{{b}} }+\frac{{e}^{−\mathrm{2}{a}} {e}^{−{nb}} }{\mathrm{2}}.\frac{{e}^{\left({n}+\mathrm{1}\right){b}} −{e}^{−\left({n}+\mathrm{1}\right){b}} }{{e}^{{b}} −{e}^{−{b}} } \\ $$$$\:=\:\:\frac{{e}^{\mathrm{2}{a}+{nb}} }{\mathrm{2}}.\frac{\mathrm{sh}\left(\left({n}+\mathrm{1}\right){b}\right)}{\mathrm{sh}{b}}+\frac{{e}^{−\left(\mathrm{2}{a}+{nb}\right)} }{\mathrm{2}}.\frac{\mathrm{sh}\left(\left({n}+\mathrm{1}\right){b}\right)}{\mathrm{sh}{b}} \\ $$$$\:=\:\:\frac{\mathrm{ch}\left(\mathrm{2}{a}+{nb}\right)\mathrm{sh}\left(\left({n}+\mathrm{1}\right){b}\right)}{\mathrm{sh}{b}} \\ $$$$ \\ $$$$\left(\mathrm{1}\right)\::\:\mathrm{S}\:=\:\frac{{n}+\mathrm{1}}{\mathrm{2}}+\:\frac{\mathrm{ch}\left(\mathrm{2}{a}+{nb}\right)\mathrm{sh}\left(\left({n}+\mathrm{1}\right){b}\right)}{\mathrm{2sh}{b}} \\ $$

Commented by puissant last updated on 23/Jul/21

$$\mathrm{jolie}\:\mathrm{prof}\:\mathrm{merci}.. \\ $$

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