Question Number 144395 by SOMEDAVONG last updated on 25/Jun/21 $$\mathrm{L}=\underset{\mathrm{n}\rightarrow+\propto} {\mathrm{lim}}\frac{\mathrm{n}+\mathrm{n}^{\mathrm{2}} +\mathrm{n}^{\mathrm{3}} +\mathrm{n}^{\mathrm{4}} +…+\mathrm{n}^{\mathrm{n}} }{\mathrm{1}^{\mathrm{n}} +\mathrm{2}^{\mathrm{n}} +\mathrm{3}^{\mathrm{n}} +\mathrm{4}^{\mathrm{n}} +…+\mathrm{n}^{\mathrm{n}} }\:=? \\ $$ Terms of Service…
Question Number 144380 by SOMEDAVONG last updated on 25/Jun/21 $$\mathrm{L}=\underset{\mathrm{n}\rightarrow+\propto} {\mathrm{lim}}\left(\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} +\mathrm{2}\left(\mathrm{1}\right)}\:+\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} +\mathrm{2n}}\:+…+\:\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} +\mathrm{2n}}\right)=? \\ $$ Answered by mathmax by abdo last updated on 25/Jun/21…
Question Number 144387 by SOMEDAVONG last updated on 25/Jun/21 $$\mathrm{A}=\left(\left(\mathrm{log}_{\mathrm{2}} \mathrm{9}\right)^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{log}_{\mathrm{2}} \left(\mathrm{log}_{\mathrm{2}} \mathrm{9}\right)}} ×\left(\sqrt{\mathrm{7}}\right)^{\frac{\mathrm{1}}{\mathrm{log}_{\mathrm{4}} \mathrm{7}}} =? \\ $$ Answered by liberty last updated on…
Question Number 144373 by help last updated on 25/Jun/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 144374 by help last updated on 25/Jun/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 78794 by naka3546 last updated on 20/Jan/20 $${f}\left({x}\:+\:\frac{\mathrm{1}}{{x}}\right)\:\:=\:\:\frac{{x}^{\mathrm{6}} \:+\:\mathrm{1}}{\mathrm{27}} \\ $$$${f}\left({x}\right)\:\:=\:\:… \\ $$ Answered by mr W last updated on 20/Jan/20 $${t}={x}+\frac{\mathrm{1}}{{x}} \\…
Question Number 144327 by SOMEDAVONG last updated on 24/Jun/21 $$\mathrm{Give}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{5}} −\mathrm{5x}^{\mathrm{4}} +\mathrm{4x}^{\mathrm{3}} −\mathrm{3x}^{\mathrm{2}} +\mathrm{2x}−\mathrm{1},\& \\ $$$$\alpha=\:\sqrt[{\mathrm{3}}]{\mathrm{2}}\left(\sqrt[{\mathrm{3}}]{\mathrm{5}+\mathrm{3}\sqrt{\mathrm{3}}}−\:\frac{\sqrt[{\mathrm{3}}]{\mathrm{2}}}{\:\sqrt[{\mathrm{3}}]{\mathrm{5}+\mathrm{3}\sqrt{\mathrm{3}}}}\right).\mathrm{Find}\:\mathrm{f}\left(\alpha\right)? \\ $$ Answered by Rasheed.Sindhi last updated on 25/Jun/21…
Question Number 13237 by Abbas-Nahi last updated on 17/May/17 Commented by RasheedSindhi last updated on 17/May/17 $$\frac{\mathrm{36}}{\left(\mathrm{n}+\mathrm{3}\right)!}−\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{1}\right)!}−\frac{\mathrm{1}}{\mathrm{n}!}=\mathrm{0} \\ $$$$\frac{\mathrm{1}}{\mathrm{n}!}\left(\frac{\mathrm{36}}{\left(\mathrm{n}+\mathrm{3}\right)\left(\mathrm{n}+\mathrm{2}\right)\left(\mathrm{n}+\mathrm{1}\right)}−\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{1}\right)}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\frac{\mathrm{1}}{\mathrm{n}!}=\mathrm{0}\:\mid\:\left(\frac{\mathrm{36}}{\left(\mathrm{n}+\mathrm{3}\right)\left(\mathrm{n}+\mathrm{2}\right)\left(\mathrm{n}+\mathrm{1}\right)}−\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{1}\right)}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\frac{\mathrm{1}}{\mathrm{n}!}=\mathrm{0}\Rightarrow\mathrm{n}\rightarrow\infty \\ $$…
Question Number 78770 by M±th+et£s last updated on 20/Jan/20 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 144307 by mohammad17 last updated on 24/Jun/21 $${lim}_{{n}\rightarrow\infty} \left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)^{{p}} \\ $$ Commented by mohammad17 last updated on 24/Jun/21 $${help}\:{me}\:{sir}\:{please} \\ $$ Answered by…