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Question Number 147820 by puissant last updated on 23/Jul/21

$$\underset{\mathrm{p}=0}{\sum ^{\mathrm{n}}}{\mathrm{sh}}^2(\mathrm{a}+\mathrm{pb})$$

Answered by Olaf_Thorendsen last updated on 23/Jul/21

$$$$$$\underset{p=0}{\sum ^n}{\mathrm{sh}}^2(\mathrm{a}+\mathrm{p}b)=\underset{p=0}{\sum ^n}({\mathrm{ch}}^2(a+pb)-1)$$$$=\underset{p=0}{\sum ^n}{\mathrm{ch}}^2(a+pb)-(n+1)$$$$=\frac{n+1}{2}+\frac{\mathrm{ch}(2a+nb)\mathrm{sh}\left((n+1)b\right)}{2\mathrm{sh}b}-(n+1)$$$$=\frac{\mathrm{ch}(2a+nb)\mathrm{sh}\left((n+1)b\right)}{2\mathrm{sh}b}-\frac{n+1}{2}$$

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