Question Number 72048 by ahmadshahhimat775@gmail.com last updated on 23/Oct/19 Answered by mind is power last updated on 23/Oct/19 $$\mathrm{a}_{\mathrm{2}} =\frac{\mathrm{6}}{\mathrm{1}+\mathrm{3}}=\frac{\mathrm{3}}{\mathrm{2}}=\frac{\mathrm{3}.\mathrm{2}^{\mathrm{0}} }{\mathrm{2}}=\frac{\mathrm{2}+\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{2}+\mathrm{2}^{\mathrm{0}} }{\mathrm{2}} \\ $$$$\mathrm{a}_{\mathrm{3}} =\frac{\mathrm{3}}{\mathrm{1}+\frac{\mathrm{3}}{\mathrm{2}}}=\frac{\mathrm{6}}{\mathrm{5}}\:=\frac{\mathrm{3}.\mathrm{2}}{\mathrm{3}.\mathrm{2}−\mathrm{1}}=\frac{\mathrm{4}+\mathrm{2}}{\mathrm{4}+\mathrm{2}−\mathrm{1}}=\frac{\mathrm{2}+\mathrm{2}^{\mathrm{2}}…
Question Number 72046 by aliesam last updated on 23/Oct/19 Commented by mathmax by abdo last updated on 26/Oct/19 $${let}\:{U}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} {nln}\left(\mathrm{1}+\left(\frac{{x}}{{n}}\right)^{\alpha} \right){dx}\:\Rightarrow{U}_{{n}} =_{\frac{{x}}{{n}}={t}} \:\:\:\int_{\mathrm{0}}…
Question Number 72047 by aliesam last updated on 23/Oct/19 Commented by mind is power last updated on 23/Oct/19 $$\mathrm{x}_{\mathrm{1}} =\mathrm{1},\mathrm{x}_{\mathrm{2}} =\mathrm{4},\mathrm{x}_{\mathrm{3}} =\mathrm{9},\mathrm{x}_{\mathrm{4}} =\mathrm{16} \\ $$$$\forall_{\mathrm{n}}…
Question Number 137582 by SOMEDAVONG last updated on 04/Apr/21 $$\mathrm{A}=\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{3}}\left(\sqrt[{\mathrm{3}}]{\mathrm{2}}−\mathrm{1}\right)\left(\sqrt[{\mathrm{3}}]{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$ Answered by Ñï= last updated on 04/Apr/21 $${A}=\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{3}}\left(\sqrt[{\mathrm{3}}]{\mathrm{2}}−\mathrm{1}\right)\left(\sqrt[{\mathrm{3}}]{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$=\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{3}}\left(\sqrt[{\mathrm{3}}]{\mathrm{4}}−\mathrm{1}\right)\left(\sqrt[{\mathrm{3}}]{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{2}} }=\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{3}}\left(\sqrt[{\mathrm{3}}]{\mathrm{4}}−\mathrm{1}\right)\left(\sqrt[{\mathrm{3}}]{\mathrm{4}}+\mathrm{2}\sqrt[{\mathrm{3}}]{\mathrm{2}}+\mathrm{1}\right)}…
Question Number 6491 by Temp last updated on 29/Jun/16 $$\mathrm{Is}\:\mathrm{it}\:\mathrm{possible}\:\mathrm{such}\:\mathrm{that}\:\mathrm{two}\:\mathrm{trancendental} \\ $$$$\mathrm{numbers}\:\mathrm{add}\:\mathrm{or}\:\mathrm{multiply}\:\mathrm{to}\:\mathrm{give}\:\mathrm{a}\:\mathrm{whole}\:\mathrm{number}? \\ $$ Commented by prakash jain last updated on 29/Jun/16 $$\pi\:\mathrm{is}\:\mathrm{transcendental}. \\ $$$$\frac{\mathrm{1}}{\pi}\:\mathrm{is}\:\mathrm{also}\:\mathrm{transcendental}.…
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Question Number 137530 by ZiYangLee last updated on 03/Apr/21 $$\mathrm{If}\:\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{sin}\:\alpha+\mathrm{sin}\:\beta\right)+\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{sin}\:\alpha−\mathrm{sin}\:\beta\right)=\frac{\pi}{\mathrm{2}} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{sin}^{\mathrm{2}} \alpha+\mathrm{sin}^{\mathrm{2}} \beta.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[\frac{\mathrm{1}}{\mathrm{2}}\right] \\ $$ Answered by mr W last updated on…
Question Number 6449 by Temp last updated on 27/Jun/16 $${x}=\mathrm{0}.\mathrm{112123123412345123456}… \\ $$$$\mathrm{decimal}\:\mathrm{pattern}:\:\mathrm{1},\:\mathrm{12},\:\mathrm{123},\:… \\ $$$$\mathrm{Can}\:\mathrm{this}\:\mathrm{be}\:\mathrm{represented}\:\mathrm{as}\:\mathrm{a}\:\mathrm{fraction}? \\ $$$$\mathrm{or}\:\mathrm{is}\:\mathrm{this}\:\mathrm{number}\:\mathrm{trancendental}? \\ $$ Commented by prakash jain last updated on…
Question Number 137523 by SOMEDAVONG last updated on 03/Apr/21 $$\mathrm{L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{x}!−\mathrm{1}}{\mathrm{x}} \\ $$ Answered by Dwaipayan Shikari last updated on 03/Apr/21 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\Gamma\left({x}+\mathrm{1}\right)−\mathrm{1}}{{x}}=\frac{\mathrm{1}−\gamma{x}−\mathrm{1}}{{x}}=−\gamma \\ $$$$\Gamma\left({x}+\mathrm{1}\right)\sim\mathrm{1}−\gamma{x}…