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Question Number 98135 by Tony Lin last updated on 11/Jun/20 $${how}\:{to}\:{split}\:{x}^{\mathrm{2}} +{xy}+{y}^{\mathrm{2}} \:{into} \\ $$$$\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\left({x}+{y}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}\left({x}−{y}\right)^{\mathrm{2}} \:? \\ $$ Commented by MJS last updated on…
Question Number 98116 by M±th+et+s last updated on 11/Jun/20 $${prove}\:{the}\:{compact}\:{form}\:{of}\:{multivariable} \\ $$$${Taylor}\:{series} \\ $$$$ \\ $$$$\left({T}\ast{B}\right)=\underset{{n}=\mathrm{1},{n}=\mathrm{2}..{n}_{{d}} =\mathrm{0}} {\overset{\infty} {\sum}}\frac{\prod_{\mathrm{i}=\mathrm{0}} ^{{n}_{{d}} } \left({x}_{\mathrm{i}} −{d}_{\mathrm{i}} \right)}{\prod_{\mathrm{i}=\mathrm{0}} ^{{n}_{{d}}…
Question Number 98112 by M±th+et+s last updated on 11/Jun/20 $$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\mathrm{2}^{{n}^{\mathrm{2}} } } \\ $$ Answered by maths mind last updated on 11/Jun/20 $${lets}\:{use}…
Question Number 98060 by M±th+et+s last updated on 11/Jun/20 $${show}\:{that} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{arctan}\left(\frac{\mathrm{2}}{{n}^{\mathrm{2}} }\right)=\frac{\mathrm{3}\pi}{\mathrm{4}} \\ $$ Answered by smridha last updated on 11/Jun/20 $${is}\:{it}\:\boldsymbol{{right}}\:\boldsymbol{{ans}}???\boldsymbol{{I}}\:{think}\:\boldsymbol{{it}}'{s}\:{just}…
Question Number 98008 by john santu last updated on 11/Jun/20 Commented by Algoritm last updated on 11/Jun/20 $$\mathrm{translate}\:? \\ $$ Terms of Service Privacy Policy…
Question Number 97956 by M±th+et+s last updated on 10/Jun/20 $${prove} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{9}{n}+\mathrm{4}}{\mathrm{3}{n}\left(\mathrm{3}{n}+\mathrm{1}\right)\left(\mathrm{3}{n}+\mathrm{2}\right)}=\frac{\mathrm{3}}{\mathrm{2}}−{ln}\left(\mathrm{3}\right) \\ $$ Answered by maths mind last updated on 13/Jun/20 $$=\underset{{n}\geqslant\mathrm{1}}…
Question Number 163408 by mathlove last updated on 06/Jan/22 $$\sqrt[{\mathrm{3}}]{\mathrm{1}+\frac{\mathrm{2}}{\mathrm{9}}\sqrt{\mathrm{21}}}+\sqrt[{\mathrm{3}}]{\mathrm{1}−\frac{\mathrm{2}}{\mathrm{9}}\sqrt{\mathrm{21}}}=? \\ $$ Answered by mr W last updated on 06/Jan/22 $$\sqrt[{\mathrm{3}}]{\mathrm{1}+\frac{\mathrm{2}}{\mathrm{9}}\sqrt{\mathrm{21}}}+\sqrt[{\mathrm{3}}]{\mathrm{1}−\frac{\mathrm{2}}{\mathrm{9}}\sqrt{\mathrm{21}}}={x} \\ $$$$\left.\mathrm{1}+\frac{\mathrm{2}}{\mathrm{9}}\sqrt{\mathrm{21}}+\mathrm{1}−\frac{\mathrm{2}}{\mathrm{9}}\sqrt{\mathrm{21}}+\mathrm{3}{x}\sqrt[{\mathrm{3}}]{\left(\mathrm{1}+\frac{\mathrm{2}}{\mathrm{9}}\sqrt{\mathrm{21}}\right)\left(\mathrm{1}−\frac{\mathrm{2}}{\mathrm{9}}\sqrt{\mathrm{21}}\right.}\right)={x}^{\mathrm{3}} \\ $$$${x}^{\mathrm{3}}…
Question Number 32328 by abdo imad last updated on 23/Mar/18 $${prove}\:{that}\:\mathrm{2}^{{n}+\mathrm{1}} \:{divide}\:\:{A}_{{n}} =\left[\left(\mathrm{1}+\sqrt{\mathrm{3}}\:\right)^{\mathrm{2}{n}+\mathrm{1}} \right]. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com