Question Number 205884 by Abdullahrussell last updated on 01/Apr/24 Commented by Frix last updated on 01/Apr/24 $${a}^{\mathrm{3}} +{b}^{\mathrm{3}} \\ $$ Answered by Rasheed.Sindhi last updated…
Question Number 205885 by hardmath last updated on 01/Apr/24 $$\mathrm{Find}:\:\:\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}}\:\sqrt[{\boldsymbol{\mathrm{n}}}]{}\begin{pmatrix}{\mathrm{2n}}\\{\:\:\mathrm{n}}\end{pmatrix}\:=\:? \\ $$ Answered by MM42 last updated on 03/Apr/24 $$\begin{pmatrix}{\mathrm{2}{n}}\\{\:\:{n}}\end{pmatrix}=\frac{\left(\mathrm{2}{n}\right)\left(\mathrm{2}{n}−\mathrm{1}\right)\left(\mathrm{2}{n}−\mathrm{2}\right)…\left({n}+\mathrm{3}\right)\left({n}+\mathrm{2}\right)\left({n}+\mathrm{1}\right)}{{n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)…\mathrm{3}×\mathrm{2}×\mathrm{1}} \\ $$$${a}=\sqrt[{{n}}]{}\begin{pmatrix}{\mathrm{2}{n}}\\{\:\:{n}}\end{pmatrix} \\ $$$$\Rightarrow{lna}=\frac{\mathrm{1}}{{n}}\left[{ln}\left(\mathrm{2}\right)+{ln}\left(\mathrm{2}+\frac{\mathrm{1}}{{n}−\mathrm{1}}\right)+{ln}\left(\mathrm{2}+\frac{\mathrm{2}}{{n}−\mathrm{2}}\right)+…+{ln}\left(\mathrm{2}+\frac{{n}−\mathrm{3}}{\mathrm{3}}\right)\left(\mathrm{2}+\frac{{n}−\mathrm{2}}{\mathrm{2}}\right)\left(\mathrm{2}+\frac{{n}−\mathrm{1}}{\mathrm{1}}\right)\right.…
Question Number 205817 by hardmath last updated on 31/Mar/24 $$\mathrm{2}^{\:\boldsymbol{\mathrm{x}}\:+\:\boldsymbol{\mathrm{log}}_{\mathrm{2}} \:\mathrm{3}} \:=\:\mathrm{12}\:\:\Rightarrow\:\:\mathrm{find}:\:\:\mathrm{x}\:=\:? \\ $$ Answered by A5T last updated on 31/Mar/24 $$\left(\mathrm{2}^{{x}} \right)\left(\mathrm{2}^{{log}_{\mathrm{2}} \mathrm{3}} \right)=\mathrm{12}\Rightarrow\left(\mathrm{2}^{{x}}…
Question Number 205772 by mr W last updated on 30/Mar/24 Answered by MM42 last updated on 30/Mar/24 $${S}_{{n}} =\left(\sqrt{\mathrm{1}×\mathrm{2}}−\mathrm{1}\right) \\ $$$$+\left(\sqrt{\mathrm{2}×\mathrm{3}}−\sqrt{\mathrm{1}×\mathrm{2}}−\mathrm{1}\right) \\ $$$$+\left(\sqrt{\mathrm{3}×\mathrm{4}}−\sqrt{\mathrm{2}×\mathrm{3}}−\mathrm{1}\right) \\ $$$$\vdots…
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Question Number 205767 by lmcp1203 last updated on 30/Mar/24 $$ \\ $$$${a},{b},{c}\:\in\Re^{+} \:\: \\ $$$${a}+{b}+{c}=\mathrm{1} \\ $$$$\:\:\:{a}^{\mathrm{2}} /\left(\mathrm{1}+{b}+{c}\right)\:+\:{b}^{\mathrm{2}} /\left(\mathrm{1}+{a}+{c}\right)\:\:+\:{c}^{\mathrm{2}} /\left(\mathrm{1}+{a}+{b}\right)\geqslant{k} \\ $$$${find}\:\:\:{k}\:{max}. \\ $$$${hint}\::\:{inequality}\:{cauchy}\:{schwarz} \\…
Question Number 205770 by hardmath last updated on 30/Mar/24 $$\mathrm{If}\:\:\mathrm{x},\mathrm{y},\mathrm{z}>\mathrm{0}\:\:\mathrm{and}\:\:\mathrm{xyz}\:=\:\mathrm{1} \\ $$$$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\frac{\left(\sqrt{\mathrm{2}}\mathrm{x}\right)^{\mathrm{2}} }{\left(\mathrm{1}+\mathrm{xz}\right)\left(\mathrm{1}+\mathrm{xy}\right)}\:+\:\frac{\left(\sqrt{\mathrm{2}}\mathrm{y}\right)^{\mathrm{2}} }{\left(\mathrm{1}+\mathrm{yz}\right)\left(\mathrm{1}+\mathrm{xy}\right)}\:+\:\frac{\left(\sqrt{\mathrm{2}}\mathrm{z}\right)^{\mathrm{2}} }{\left(\mathrm{1}+\mathrm{xz}\right)\left(\mathrm{1}+\mathrm{yz}\right)}\:\geqslant\:\frac{\mathrm{3}}{\mathrm{2}} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 205746 by Satyam1234 last updated on 29/Mar/24 Answered by A5T last updated on 29/Mar/24 $${C}\neq{J}+\mathrm{2}\Rightarrow{C}=\mathrm{3}{F} \\ $$$${A}\geqslant\mathrm{1};{A}=\mathrm{1}\Rightarrow{B}=\mathrm{3}\Rightarrow{I}=\mathrm{7}\Rightarrow{H}=\mathrm{4}\Rightarrow{A}=\mathrm{8}\:{X} \\ $$$${A}=\mathrm{2}\Rightarrow{B}=\mathrm{6}\Rightarrow{E}=\mathrm{2}\:{X} \\ $$$${A}=\mathrm{3}\Rightarrow{B}=\mathrm{9}\:{X}\left(\mathrm{6}\Rightarrow{B}<\mathrm{5}\:{or}\:{A}\geqslant\mathrm{4}\right)\:{or}\:{G}=\mathrm{0} \\ $$$$\:\left({G}=\mathrm{0}\Rightarrow{I}=\mathrm{3}\Rightarrow{D}=\mathrm{6}\Rightarrow{E}=\mathrm{18}\right){X}…