Question Number 65739 by Masumsiddiqui399@gmail.com last updated on 03/Aug/19 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 65674 by mathmax by abdo last updated on 01/Aug/19 $${find}\:{the}\:{value}\:{of}\:\sum_{{n}=\mathrm{2}} ^{\infty} \:\frac{{n}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} \left({n}−\mathrm{1}\right)^{\mathrm{3}} } \\ $$ Commented by mathmax by abdo last updated…
Question Number 65677 by mathmax by abdo last updated on 01/Aug/19 $${let}\:{S}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{\:\sqrt{{k}^{\mathrm{2}} +{k}}} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{a}\:{equivalent}\:{of}\:{S}_{{n}} \:{when}\:{n}\rightarrow+\infty \\ $$$$\left.\mathrm{2}\right){prove}\:{that}\:\left({S}_{{n}} \right)\:{is}\:{convergent}. \\ $$ Terms…
Question Number 65673 by mathmax by abdo last updated on 01/Aug/19 $${find}\:\sum_{{n}=\mathrm{2}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left({n}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} } \\ $$ Commented by mathmax by abdo last…
Question Number 65664 by mathmax by abdo last updated on 01/Aug/19 $${solve}\:\frac{\sqrt{\mathrm{1}−{x}}−\sqrt{\mathrm{2}{x}+\mathrm{1}}}{\:\sqrt{\mathrm{1}−{x}}+\sqrt{\mathrm{2}{x}+\mathrm{1}}}\:=\frac{{x}+\mathrm{1}}{\mathrm{3}} \\ $$ Answered by MJS last updated on 01/Aug/19 $$\frac{\sqrt{\mathrm{1}−{x}}−\sqrt{\mathrm{2}{x}+\mathrm{1}}}{\:\sqrt{\mathrm{1}−{x}}+\sqrt{\mathrm{2}{x}+\mathrm{1}}}=\frac{\left(\sqrt{\mathrm{1}−{x}}−\sqrt{\mathrm{2}{x}+\mathrm{1}}\right)^{\mathrm{2}} }{\left(\sqrt{\mathrm{1}−{x}}+\sqrt{\mathrm{2}{x}+\mathrm{1}}\right)\left(\sqrt{\mathrm{1}−{x}}−\sqrt{\mathrm{2}{x}+\mathrm{1}}\right)}= \\ $$$$=\frac{{x}+\mathrm{2}−\mathrm{2}\sqrt{\mathrm{1}−{x}}\sqrt{\mathrm{2}{x}+\mathrm{1}}}{−\mathrm{3}{x}}=−\frac{{x}+\mathrm{2}}{\mathrm{3}{x}}+\frac{\mathrm{2}\sqrt{\mathrm{1}−{x}}\sqrt{\mathrm{2}{x}+\mathrm{1}}}{\mathrm{3}{x}}…
Question Number 144001 by mnjuly1970 last updated on 20/Jun/21 Answered by Dwaipayan Shikari last updated on 20/Jun/21 $$\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)^{\frac{\mathrm{1}}{\mathrm{1}}} \left(\mathrm{1}+\frac{\mathrm{2}}{{n}}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{1}+\frac{\mathrm{3}}{{n}}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} …={y} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{{n}}\underset{{k}=\mathrm{1}} {\overset{{n}}…
Question Number 143987 by liberty last updated on 20/Jun/21 $${If}\:{f}\left({x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{6}\right)+{f}\left({x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{4}\right)=\mathrm{2}{x} \\ $$$$\forall{x}\in{R}\:{th}\mathrm{e}{n}\:{f}\left(−\mathrm{3}\right)+{f}\left(\mathrm{9}\right)−\mathrm{5}{f}\left(\mathrm{1}\right)=? \\ $$ Answered by mitica last updated on 20/Jun/21 $${x}=\mathrm{1}\Rightarrow{f}\left(\mathrm{1}\right)+{f}\left(\mathrm{1}\right)=\mathrm{2}\Rightarrow{f}\left(\mathrm{1}\right)=\mathrm{1} \\…
Question Number 12855 by kunalshukla95040 last updated on 04/May/17 $${f}\left({x}\right)=\left[\mathrm{1}−\left({x}−\mathrm{3}\right)^{\mathrm{4}} \right]^{\mathrm{1}/\mathrm{7}} \\ $$$${find}\:{f}^{−\mathrm{1}} \left({x}\right). \\ $$ Answered by linkelly0615 last updated on 04/May/17 $${f}\left({x}\right)=\left[\mathrm{1}−\left({x}−\mathrm{3}\right)^{\mathrm{4}} \right]^{\mathrm{1}/\mathrm{7}}…
Question Number 143860 by mathmax by abdo last updated on 19/Jun/21 $$\mathrm{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\mathrm{x}\:\mathrm{e}^{−\mathrm{x}^{\mathrm{2}} } \mathrm{log}\left(\mathrm{1}+\mathrm{e}^{\mathrm{x}} \right)\mathrm{dx} \\ $$ Answered by mathmax by abdo last…
Question Number 143846 by mathmax by abdo last updated on 18/Jun/21 $$\mathrm{find}\:\mathrm{the}\:\mathrm{sum}\:\:\frac{\left(\mathrm{0}!\right)^{\mathrm{2}} }{\mathrm{1}!}\:+\frac{\left(\mathrm{1}!\right)^{\mathrm{2}} }{\mathrm{3}!}\:+\frac{\left(\mathrm{2}!\right)^{\mathrm{2}} }{\mathrm{5}!}\:+….. \\ $$ Answered by Ar Brandon last updated on 18/Jun/21…