Question Number 63858 by gunawan last updated on 10/Jul/19 $$\mathrm{if}\:\mathrm{a}_{\mathrm{1}} ,\:\mathrm{a}_{\mathrm{2}} ,\:\mathrm{a}_{\mathrm{3}} ,\:\mathrm{a}_{\mathrm{4}} \:\mathrm{are}\:\mathrm{the}\:\mathrm{coefficient} \\ $$$$\mathrm{of}\:\mathrm{any}\:\mathrm{four}\:\mathrm{four}\:\mathrm{consecutive} \\ $$$$\mathrm{terms}\:\mathrm{in}\:\mathrm{the}\:\mathrm{expansion}\:\mathrm{of}\:\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{n}} \\ $$$$\mathrm{then}\:\frac{\mathrm{a}_{\mathrm{1}} }{\mathrm{a}_{\mathrm{2}} +\mathrm{a}_{\mathrm{1}} }+\frac{\mathrm{a}_{\mathrm{3}} }{\mathrm{a}_{\mathrm{3}} +\mathrm{a}_{\mathrm{4}}…
Question Number 63857 by ajfour last updated on 10/Jul/19 Commented by ajfour last updated on 10/Jul/19 $${perfect}!\:{Sir}. \\ $$ Commented by mr W last updated…
Question Number 129388 by MathSh last updated on 15/Jan/21 $${Solve}\:{the}\:{equation}: \\ $$$${y}''−\mathrm{2}{y}'+\mathrm{2}{y}={e}^{{x}} +{xcosx} \\ $$ Answered by mathmax by abdo last updated on 16/Jan/21 $$\mathrm{h}\rightarrow\mathrm{r}^{\mathrm{2}}…
Question Number 63852 by aliesam last updated on 10/Jul/19 $${prove}\:{that} \\ $$$$ \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {arctan}\left({x}\right)\:{cot}\left(\frac{\pi{x}}{\mathrm{2}}\right)\:{dx}\:=\:\frac{\mathrm{3}\:{ln}^{\mathrm{2}} \left(\mathrm{2}\right)}{\mathrm{2}\pi}+\frac{{ln}\pi\:{ln}\mathrm{2}}{\pi}+\int_{\mathrm{0}} ^{\infty} \frac{{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{{e}^{\mathrm{2}\pi{x}} +\mathrm{1}}\:{dx} \\ $$ Terms of…
Question Number 129387 by Study last updated on 15/Jan/21 $${li}\underset{{x}\rightarrow\mathrm{0}} {{m}}\left({cosx}\right)^{{logx}} =??? \\ $$ Commented by Study last updated on 15/Jan/21 $${who}\:{will}\:{solve}? \\ $$ Answered…
Question Number 129385 by Adel last updated on 15/Jan/21 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{\:\sqrt{\mathrm{n}\:}}\frac{\:}{\:\sqrt{\mathrm{n}+\mathrm{1}}}+\frac{\:}{\:\sqrt{\mathrm{n}}}\frac{\:\mathrm{1}}{\:\sqrt{\mathrm{n}+\mathrm{2}}}\frac{\:}{\:}+\frac{\:}{\:\sqrt{\boldsymbol{\mathrm{n}}}}\frac{\:}{\:\sqrt{\boldsymbol{\mathrm{n}}+\mathrm{3}}}\frac{\mathrm{1}}{\:}\frac{\:}{\:}+…………..+\frac{\:}{\:\sqrt{\mathrm{n}}}\frac{\mathrm{1}}{\:\sqrt{\mathrm{n}+\mathrm{n}}}\frac{\:}{\:}=? \\ $$$$\:{what}\:{answer} \\ $$ Answered by Dwaipayan Shikari last updated on 15/Jan/21 $$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\underset{{k}=\mathrm{1}}…
Question Number 129383 by mathmax by abdo last updated on 15/Jan/21 $$\mathrm{find}\:\mathrm{the}\:\mathrm{sequence}\:\mathrm{u}_{\mathrm{n}} \:\mathrm{wish}\:\mathrm{verify}\:\mathrm{u}_{\mathrm{n}−\mathrm{p}} +\mathrm{u}_{\mathrm{n}+\mathrm{p}} =\frac{\left(−\mathrm{1}\right)^{\mathrm{p}} }{\mathrm{n}+\mathrm{p}} \\ $$$$\mathrm{for}\:\mathrm{all}\:\mathrm{n}\geqslant\mathrm{p} \\ $$ Terms of Service Privacy Policy…
Question Number 129380 by mohammad17 last updated on 15/Jan/21 Answered by physicstutes last updated on 15/Jan/21 $$\left\{{a}_{{n}} \right\}\:=\:{n}^{\mathrm{2}} \\ $$$${a}_{{n}+\mathrm{1}} \:=\:\left({n}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\:{a}_{{n}+\mathrm{1}} −{a}_{{n}} \:=\:\left({n}+\mathrm{1}\right)^{\mathrm{2}}…
Question Number 63845 by gunawan last updated on 10/Jul/19 $$\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}\:^{{n}} {C}_{{r}} \:\frac{\mathrm{1}+{r}\:\mathrm{log}_{{e}} \:\mathrm{10}}{\left(\mathrm{1}+\:\mathrm{log}_{{e}} \:\mathrm{10}^{{n}} \right)^{{r}} }=… \\ $$ Commented by gunawan last updated…
Question Number 63844 by mmkkmm000m last updated on 10/Jul/19 $$\int\left(\mathrm{1}+\mathrm{4}{x}+{x}^{\mathrm{2}} \right)^{{m}} {dx} \\ $$ Commented by mathmax by abdo last updated on 10/Jul/19 $${let}\:{A}_{{m}} =\int\:\left({x}^{\mathrm{2}}…