Question Number 103881 by bemath last updated on 18/Jul/20 $$\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\frac{\left(\mathrm{2}{n}−\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{4}{n}^{\mathrm{2}} }\:? \\ $$ Answered by bramlex last updated on 18/Jul/20 $${use}\:{Euler}\:{infinity}\:{product}\: \\ $$$$\frac{\mathrm{sin}\:{x}}{{x}}\:=\:\underset{{n}=\mathrm{1}}…
Question Number 38332 by ajfour last updated on 24/Jun/18 $$\:\frac{\mathrm{3}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:−\sqrt{\frac{\mathrm{3}+\sqrt{\mathrm{5}}}{\mathrm{2}}}\:=\:?! \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 24/Jun/18 $${x}=\frac{\mathrm{3}+\sqrt{\mathrm{5}\:}}{\mathrm{2}} \\ $$$$=\frac{\mathrm{6}+\mathrm{2}\sqrt{\mathrm{5}}\:}{\mathrm{4}} \\ $$$$=\left(\frac{\sqrt{\mathrm{5}}\:+\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} =\pm\frac{\sqrt{\mathrm{5}}\:+\mathrm{1}}{\mathrm{2}}…
Question Number 103832 by bemath last updated on 17/Jul/20 $${p}\left({x}\right)\:=\:{x}^{\mathrm{4}} +{ax}^{\mathrm{3}} +{bx}^{\mathrm{2}} +{cx}\:+{d} \\ $$$${if}\:{p}\left(\mathrm{1}\right)=\mathrm{10},{p}\left(\mathrm{2}\right)=\mathrm{20}\:{and} \\ $$$${p}\left(\mathrm{3}\right)=\mathrm{30}\:.\:{find}\:\frac{{p}\left(\mathrm{12}\right)+{p}\left(−\mathrm{8}\right)}{\mathrm{10}} \\ $$ Answered by floor(10²Eta[1]) last updated on…
Question Number 103796 by bemath last updated on 17/Jul/20 $$\underset{{k}\:=\:\mathrm{1}} {\overset{{n}} {\sum}}{k}^{\mathrm{5}} \:=\:? \\ $$ Commented by mr W last updated on 17/Jul/20 Commented by…
Question Number 103780 by bobhans last updated on 17/Jul/20 $${the}\:{third},\:{sixth}\:{and}\:{seventh}\:{terms}\:{of}\:{a} \\ $$$${geometric}\:{progression}\:\left({whose}\:{common}\right. \\ $$$$\left.{ratio}\:{is}\:{neither}\:\mathrm{0}\:{nor}\:\mathrm{1}\:\right)\:{are}\:{in} \\ $$$${arithmetic}\:{progression}\:.\:{prove}\:{that}\:{the} \\ $$$${sum}\:{of}\:{the}\:{first}\:{three}\:{terms}\:{is}\:{equal}\:{to} \\ $$$${fourth}\:. \\ $$ Answered by bemath…
Question Number 103755 by bramlex last updated on 17/Jul/20 $$\underset{{n}\:=\:\mathrm{3}} {\overset{\infty} {\prod}}\left(\mathrm{1}−\mathrm{tan}\:^{\mathrm{4}} \left(\frac{\pi}{\mathrm{2}^{{n}} }\right)\right)\:=\:? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 103700 by anonymous last updated on 16/Jul/20 Commented by anonymous last updated on 16/Jul/20 $${plzz}\:{help}\:{me} \\ $$ Answered by bobhans last updated on…
Question Number 103686 by Jamshidbek2311 last updated on 16/Jul/20 $${x}^{{x}^{\mathrm{6}} } =\sqrt{\mathrm{2}}\:^{\sqrt{\mathrm{2}}} \:{x}=? \\ $$ Answered by Dwaipayan Shikari last updated on 16/Jul/20 $${x}^{\mathrm{6}} {logx}=\sqrt{\mathrm{2}}.\frac{\mathrm{1}}{\mathrm{2}}{log}\mathrm{2}…
Question Number 103673 by bobhans last updated on 16/Jul/20 $$\underset{{k}=\mathrm{1}} {\overset{\mathrm{4095}} {\sum}}\frac{\mathrm{1}}{\left(\sqrt{{k}}+\sqrt{{k}+\mathrm{1}}\right)\left(\sqrt[{\mathrm{4}}]{{k}}+\sqrt[{\mathrm{4}}]{{k}+\mathrm{1}}\right)}\:? \\ $$ Answered by bramlex last updated on 16/Jul/20 $$\frac{\mathrm{1}}{\left(\sqrt{\mathrm{k}}+\sqrt{\mathrm{k}+\mathrm{1}}\right)\left(\sqrt[{\mathrm{4}}]{\mathrm{k}}+\sqrt[{\mathrm{4}}]{\mathrm{k}+\mathrm{1}}\right)}\:×\frac{\sqrt[{\mathrm{4}}]{\mathrm{k}+\mathrm{1}}−\sqrt[{\mathrm{4}}]{\mathrm{k}}}{\:\sqrt[{\mathrm{4}}]{\mathrm{k}+\mathrm{1}}−\sqrt[{\mathrm{4}}]{\mathrm{k}}}= \\ $$$$\frac{\sqrt[{\mathrm{4}}]{\mathrm{k}+\mathrm{1}}−\sqrt[{\mathrm{4}}]{\mathrm{k}}}{\left(\sqrt{\mathrm{k}+\mathrm{1}}+\sqrt{\mathrm{k}}\right)\left(\sqrt{\mathrm{k}+\mathrm{1}}−\sqrt{\mathrm{k}}\right)}\:=\:\sqrt[{\mathrm{4}}]{\mathrm{k}+\mathrm{1}}−\sqrt[{\mathrm{4}}]{\mathrm{k}} \\…
Question Number 103670 by bobhans last updated on 16/Jul/20 $${Given}\:{b}_{{n}} \:=\:\mathrm{3}.\mathrm{2}^{{n}} \:{is}\:{a}\:{GP}\:.\:{find}\:{the}\:{value} \\ $$$${of}\:\frac{\mathrm{1}}{{b}_{\mathrm{1}} }+\frac{\mathrm{1}}{{b}_{\mathrm{2}} }+\frac{\mathrm{1}}{{b}_{\mathrm{3}} }+…+\frac{\mathrm{1}}{{b}_{\mathrm{10}} }\:?\: \\ $$ Answered by bramlex last updated…