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1-1-2-2-1-2-1-1-3-2-2-2-1-2-2-2-1-3-5-2-3-1-2-3-3-1-4-2pi-3-1-4-2-

Question Number 136757 by Dwaipayan Shikari last updated on 25/Mar/21 $$\mathrm{1}+\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \frac{\mathrm{1}}{\mathrm{2}.\mathrm{1}!}+\left(\frac{\mathrm{1}.\mathrm{3}}{\mathrm{2}^{\mathrm{2}} }\right)^{\mathrm{2}} \frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} .\mathrm{2}!}+\left(\frac{\mathrm{1}.\mathrm{3}.\mathrm{5}}{\mathrm{2}^{\mathrm{3}} }\right).\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} .\mathrm{3}!}+….=\left(\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)}{\left(\mathrm{2}\pi^{\mathrm{3}} \right)^{\mathrm{1}/\mathrm{4}} }\right)^{\mathrm{2}} \\ $$ Answered by mindispower last…

Question-136760

Question Number 136760 by I want to learn more last updated on 25/Mar/21 Answered by mr W last updated on 25/Mar/21 $$\frac{{BC}}{\mathrm{sin}\:\angle{BDC}}=\frac{{CD}}{\mathrm{sin}\:\mathrm{30}°}\:\:\:\:\:…\left({i}\right) \\ $$$$\frac{\mathrm{sin}\:\angle{BDA}}{{AB}}=\frac{\mathrm{sin}\:\mathrm{45}°}{{DA}}\:\:\:…\left({ii}\right) \\…

f-n-x-x-1-x-2x-2-x-nx-n-x-n-N-f-n-x-f-n-1-x-nx-n-x-f-1-x-x-1-x-f-n-1-x-f-n-x-nx-n-x-n-Z-f-0-x-f-1-x-x-1-x

Question Number 5585 by 123456 last updated on 21/May/16 $${f}_{{n}} \left({x}\right)=\frac{{x}}{\mathrm{1}−{x}}+\frac{\mathrm{2}{x}}{\mathrm{2}−{x}}+…+\frac{{nx}}{{n}−{x}} \\ $$$${n}\in\mathbb{N}^{\ast} \\ $$$$−−−−−−−−−−−−−−−− \\ $$$${f}_{{n}} \left({x}\right)={f}_{{n}−\mathrm{1}} \left({x}\right)+\frac{{nx}}{{n}−{x}} \\ $$$${f}_{\mathrm{1}} \left({x}\right)=\frac{{x}}{\mathrm{1}−{x}} \\ $$$${f}_{{n}−\mathrm{1}} \left({x}\right)={f}_{{n}}…