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p-0-n-sh-2-a-pb-




Question Number 147820 by puissant last updated on 23/Jul/21
Σ_(p=0) ^n sh^2 (a+pb)
$$\underset{\mathrm{p}=\mathrm{0}} {\overset{\mathrm{n}} {\sum}}\mathrm{sh}^{\mathrm{2}} \left(\mathrm{a}+\mathrm{pb}\right) \\ $$
Answered by Olaf_Thorendsen last updated on 23/Jul/21
  Σ_(p=0) ^n sh^2 (a+pb) = Σ_(p=0) ^n (ch^2 (a+pb)−1)  = Σ_(p=0) ^n ch^2 (a+pb)−(n+1)  = ((n+1)/2)+ ((ch(2a+nb)sh((n+1)b))/(2shb))−(n+1)  = ((ch(2a+nb)sh((n+1)b))/(2shb))−((n+1)/2)
$$ \\ $$$$\underset{{p}=\mathrm{0}} {\overset{{n}} {\sum}}\mathrm{sh}^{\mathrm{2}} \left(\mathrm{a}+\mathrm{p}{b}\right)\:=\:\underset{{p}=\mathrm{0}} {\overset{{n}} {\sum}}\left(\mathrm{ch}^{\mathrm{2}} \left({a}+{pb}\right)−\mathrm{1}\right) \\ $$$$=\:\underset{{p}=\mathrm{0}} {\overset{{n}} {\sum}}\mathrm{ch}^{\mathrm{2}} \left({a}+{pb}\right)−\left({n}+\mathrm{1}\right) \\ $$$$=\:\frac{{n}+\mathrm{1}}{\mathrm{2}}+\:\frac{\mathrm{ch}\left(\mathrm{2}{a}+{nb}\right)\mathrm{sh}\left(\left({n}+\mathrm{1}\right){b}\right)}{\mathrm{2sh}{b}}−\left({n}+\mathrm{1}\right) \\ $$$$=\:\frac{\mathrm{ch}\left(\mathrm{2}{a}+{nb}\right)\mathrm{sh}\left(\left({n}+\mathrm{1}\right){b}\right)}{\mathrm{2sh}{b}}−\frac{{n}+\mathrm{1}}{\mathrm{2}} \\ $$

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