Question Number 211262 by mathlove last updated on 03/Sep/24
Answered by mahdipoor last updated on 03/Sep/24
$${f}\left({m}\right)+{f}\left(\mathrm{1}−{m}\right)=\frac{{a}^{{m}−\mathrm{1}} \:}{{a}^{{m}} +{b}}+\frac{{a}^{−{m}} }{{a}^{\mathrm{1}−{m}} +{b}}= \\ $$$$\frac{{a}^{{m}−\mathrm{1}} \left({a}^{\mathrm{1}−{m}} +{b}\right)+{a}^{−{m}} \left({a}^{{m}} +{b}\right)}{\left({a}^{{m}} +{b}\right)\left({a}^{\mathrm{1}−{m}} +{b}\right)}=\frac{\mathrm{1}+{a}^{{m}−\mathrm{1}} {b}+\mathrm{1}+{a}^{−{m}} {b}}{{a}+{b}^{\mathrm{2}} +{b}\left({a}^{{m}} +{a}^{\mathrm{1}−{m}} \right)} \\ $$$$\Rightarrow\:{if}\:\:\:{a}={b}^{\mathrm{2}} \:\:\:\:\:{b}=\mathrm{22}\:\:\Rightarrow \\ $$$$=\frac{\mathrm{2}+{b}^{\mathrm{2}{m}−\mathrm{1}} +{b}^{\mathrm{1}−\mathrm{2}{m}} }{\mathrm{2}{b}^{\mathrm{2}} +{b}^{\mathrm{2}{m}+\mathrm{1}} +{b}^{\mathrm{3}−\mathrm{2}{m}} }=\frac{\mathrm{1}}{{b}^{\mathrm{2}} } \\ $$$${f}\left(\frac{\mathrm{1}}{\mathrm{45}}\right)+…+{f}\left(\frac{\mathrm{44}}{\mathrm{45}}\right)=\underset{{m}=\mathrm{1}} {\overset{\mathrm{22}} {\sum}}\left({f}\left(\frac{{m}}{\mathrm{45}}\right)+{f}\left(\mathrm{1}−\frac{{m}}{\mathrm{45}}\right)\right) \\ $$$$=\mathrm{22}×\frac{\mathrm{1}}{{b}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{22}} \\ $$
Commented by mathlove last updated on 04/Sep/24
$${thanks} \\ $$