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find-the-largest-coeeficient-in-3x-2-3-

Question Number 113794 by aurpeyz last updated on 15/Sep/20 $$\mathrm{find}\:\mathrm{the}\:\mathrm{largest}\:\mathrm{coeeficient}\:\mathrm{in}\:\left(\mathrm{3x}−\mathrm{2}\right)^{\mathrm{3}} \\ $$ Answered by mr W last updated on 15/Sep/20 $$\left(\mathrm{3}{x}−\mathrm{2}\right)^{\mathrm{3}} =\left(\mathrm{3}{x}\right)^{\mathrm{3}} +\mathrm{3}\left(\mathrm{3}{x}\right)^{\mathrm{2}} \left(−\mathrm{2}\right)+\mathrm{3}\left(\mathrm{3}{x}\right)\left(−\mathrm{2}\right)^{\mathrm{2}} +\left(−\mathrm{2}\right)^{\mathrm{3}}…

n-1-3-n-n-3-

Question Number 113789 by Khalmohmmad last updated on 15/Sep/20 $$\underset{\mathrm{n}=\mathrm{1}} {\overset{\propto} {\sum}}\frac{\mathrm{3}}{\mathrm{n}\left(\mathrm{n}+\mathrm{3}\right)}=? \\ $$ Answered by aleks041103 last updated on 15/Sep/20 $$\frac{\mathrm{3}}{{n}\left({n}+\mathrm{3}\right)}=\frac{\left({n}+\mathrm{3}\right)−{n}}{{n}\left({n}+\mathrm{3}\right)}=\frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{{n}+\mathrm{3}} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty}…

Question-113771

Question Number 113771 by mohammad17 last updated on 15/Sep/20 Answered by Dwaipayan Shikari last updated on 15/Sep/20 $$\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)−\mathrm{2}{xy}\frac{{dy}}{{dx}}=\mathrm{0} \\ $$$$\mathrm{2}{vx}^{\mathrm{2}} \frac{{dy}}{{dx}}=\left({x}^{\mathrm{2}} +{v}^{\mathrm{2}} {x}^{\mathrm{2}}…

1-i-4i-

Question Number 48227 by gunawan last updated on 21/Nov/18 $$\left(\mathrm{1}−{i}\right)^{\mathrm{4}{i}} =.. \\ $$ Answered by Smail last updated on 21/Nov/18 $$\left(\mathrm{1}−{i}\right)^{\mathrm{4}{i}} =\left(\sqrt{\mathrm{2}}\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}−{i}\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)\right)^{\mathrm{4}{i}} =\left(\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}}} \right)^{\mathrm{4}{i}} \left({e}^{−\frac{{i}\pi}{\mathrm{4}}}…

calculate-log-1-3-i-2-

Question Number 48226 by gunawan last updated on 21/Nov/18 $$\mathrm{calculate} \\ $$$$\mathrm{log}\left(−\mathrm{1}+\sqrt{\mathrm{3}}\:\mathrm{i}\right)^{\mathrm{2}} \\ $$ Answered by Smail last updated on 21/Nov/18 $${ln}\left(\left(−\mathrm{1}+\sqrt{\mathrm{3}}{i}\right)^{\mathrm{2}} \right)=\mathrm{2}{ln}\left(\mathrm{2}\left(−\frac{\mathrm{1}}{\mathrm{2}}+{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\right) \\ $$$$=\mathrm{2}{ln}\mathrm{2}+\mathrm{2}{ln}\left({cos}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right)+{isin}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right)\right)…