Question Number 128845 by Dwaipayan Shikari last updated on 10/Jan/21 | ||
$$\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{1}.\mathrm{4}}{\left(\mathrm{5}.\mathrm{1}!\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}}.\frac{\mathrm{1}.\mathrm{4}.\mathrm{6}.\mathrm{9}}{\left(\mathrm{5}^{\mathrm{2}} .\mathrm{2}!\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{4}}.\frac{\mathrm{1}.\mathrm{4}.\mathrm{6}.\mathrm{9}.\mathrm{11}.\mathrm{14}}{\left(\mathrm{5}^{\mathrm{3}} .\mathrm{3}!\right)^{\mathrm{2}} }+...=\frac{{b}^{\mathrm{2}} \sqrt{\frac{{b}−\sqrt{{b}}}{\mathrm{2}}}}{{a}\pi} \\ $$$${Find}\:\mathrm{5}{a}−\mathrm{8}{b} \\ $$ | ||
Answered by mindispower last updated on 10/Jan/21 | ||
$$=\mathrm{1}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left(\mathrm{1}+\mathrm{5}{k}\right)\left(\mathrm{4}+\mathrm{5}{k}\right)}{\left({n}+\mathrm{1}\right).\left(\mathrm{5}^{{n}} .{n}!\right)^{\mathrm{2}} } \\ $$$$=\mathrm{1}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{5}^{\mathrm{2}{n}} \underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left({k}+\frac{\mathrm{1}}{\mathrm{5}}\right).\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left({k}+\frac{\mathrm{4}}{\mathrm{5}}\right)}{\mathrm{5}^{\mathrm{2}{n}} .\left({n}+\mathrm{1}\right)!.{n}!} \\ $$$$=\mathrm{1}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(\frac{\mathrm{1}}{\mathrm{5}}\right)_{{n}} .\left(\frac{\mathrm{4}}{\mathrm{5}}\right)_{{n}} }{\left(\mathrm{2}\right)_{{n}} }.\frac{\left(\mathrm{1}\right)^{{n}} }{{n}!} \\ $$$$=_{\mathrm{2}} {F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{5}},\frac{\mathrm{4}}{\mathrm{5}};\mathrm{2};\left[\mathrm{1}\right]\right)=\frac{\Gamma\left(\mathrm{2}−\mathrm{1}\right)\Gamma\left(\mathrm{2}\right)}{\Gamma\left(\mathrm{2}−\frac{\mathrm{1}}{\mathrm{5}}\right)\Gamma\left(\mathrm{2}−\frac{\mathrm{4}}{\mathrm{5}}\right)}=\frac{\mathrm{1}}{\Gamma\left(\frac{\mathrm{9}}{\mathrm{5}}\right)\Gamma\left(\frac{\mathrm{6}}{\mathrm{5}}\right)} \\ $$$$=\Gamma\left(\frac{\mathrm{9}}{\mathrm{5}}\right)=\frac{\mathrm{4}}{\mathrm{5}}\Gamma\left(\frac{\mathrm{4}}{\mathrm{5}}\right) \\ $$$$\Gamma\left(\frac{\mathrm{6}}{\mathrm{5}}\right)=\frac{\mathrm{1}}{\mathrm{5}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{5}}\right) \\ $$$${S}=\frac{\mathrm{25}}{\mathrm{4}\Gamma\left(\frac{\mathrm{1}}{\mathrm{5}}\right)\Gamma\left(\frac{\mathrm{4}}{\mathrm{5}}\right)}=\frac{\mathrm{25}}{\mathrm{4}\pi}{sin}\left(\frac{\pi}{\mathrm{5}}\right)=\frac{\mathrm{25}}{\mathrm{4}\pi}.\frac{\mathrm{1}}{\mathrm{4}}.\sqrt{\mathrm{10}−\mathrm{2}\sqrt{\mathrm{5}}} \\ $$$$=\frac{\mathrm{25}}{\mathrm{8}\pi}.\sqrt{\frac{\mathrm{5}−\sqrt{\mathrm{5}}}{\mathrm{2}}},{b}=\mathrm{5},{a}=\mathrm{8} \\ $$$$\mathrm{5}{a}−\mathrm{8}{b}=\mathrm{0} \\ $$$$ \\ $$ | ||