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