Question Number 6350 by sanusihammed last updated on 24/Jun/16 $${Prove}\:{by}\:{matimatical}\:{induction} \\ $$$$\mathrm{1}\:+\:\mathrm{8}\:+\:\mathrm{27}\:+\:……\:+\:{n}^{\mathrm{3}} \:\:=\:\:\left[\frac{{n}\left({n}\:+\:\mathrm{1}\right)}{\mathrm{2}}\right]^{\mathrm{2}} \\ $$ Answered by prakash jain last updated on 24/Jun/16 $${n}=\mathrm{1} \\…
Question Number 137415 by mr W last updated on 02/Apr/21 $${solve} \\ $$$${x}+\sqrt{{x}\left({x}+\mathrm{1}\right)}+\sqrt{{x}\left({x}+\mathrm{2}\right)}+\sqrt{\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)}=\mathrm{2} \\ $$ Commented by MJS_new last updated on 03/Apr/21 $${x}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{24}} \\…
Question Number 6329 by Rasheed Soomro last updated on 24/Jun/16 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{n}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{the}\:\mathrm{following}\:\mathrm{series} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}+\sqrt{\mathrm{x}}}+\frac{\mathrm{1}}{\mathrm{1}−\mathrm{x}}+\frac{\mathrm{1}}{\mathrm{1}−\sqrt{\mathrm{x}}}+…. \\ $$ Commented by FilupSmith last updated on 24/Jun/16 $${s}\mathrm{equence}\:{x},\:\sqrt{{x}},\:{x}…\:\mathrm{can}\:\mathrm{be}\:\mathrm{given}\:\mathrm{by}: \\ $$$$\frac{{x}}{\mathrm{2}}\left(\mathrm{1}−\left(−\mathrm{1}\right)^{{n}+\mathrm{1}}…
Question Number 6328 by Rasheed Soomro last updated on 24/Jun/16 $$\mathrm{Sum}\:\mathrm{0}.\mathrm{7}+\mathrm{0}.\mathrm{71}+\mathrm{0}.\mathrm{72}+….\mathrm{to}\:\mathrm{100}\:\mathrm{terms}. \\ $$ Commented by FilupSmith last updated on 24/Jun/16 $$\mathrm{difference}\:{d}\:=\:\mathrm{0}.\mathrm{01} \\ $$$$\mathrm{arithmetic}\:\mathrm{series}: \\ $$$$\therefore\:{S}=\frac{{n}}{\mathrm{2}}\left(\mathrm{2}{a}+\left({n}−\mathrm{1}\right){d}\right)…
Question Number 6317 by sanusihammed last updated on 23/Jun/16 $${Solve}\:{the}\:{system}\:{of}\:{equation}\: \\ $$$$ \\ $$$${x}\:+\:{y}\:−\:{z}\:=\:\mathrm{9}\:\:\:……….\:{equation}\:\left({i}\right) \\ $$$${x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}\:} +\:{z}^{\mathrm{2}\:} =\:\mathrm{99}\:\:………..\:{equation}\:\left({ii}\right) \\ $$$${y}^{\mathrm{2}} \:=\:{xz}\:\:\:………..\:{equation}\left({iii}\right) \\ $$ Commented…
Question Number 137379 by oooooooo last updated on 02/Apr/21 Answered by Ar Brandon last updated on 02/Apr/21 $$\mathrm{I}=\int_{\mathrm{0}} ^{\frac{\mathrm{a}}{\mathrm{2}}} \sqrt{\frac{\mathrm{x}}{\mathrm{a}−\mathrm{x}}}\mathrm{dx}=\int_{\mathrm{0}} ^{\frac{\mathrm{a}}{\mathrm{2}}} \frac{\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{2}}} }{\:\sqrt{\mathrm{a}−\mathrm{x}}}\mathrm{dx} \\ $$$$\mathrm{u}=\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{2}}}…
Question Number 137376 by benjo_mathlover last updated on 02/Apr/21 $$\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{9}}+\frac{\mathrm{1}}{\mathrm{15}}+\frac{\mathrm{1}}{\mathrm{25}}+…+\frac{\mathrm{1}}{\mathrm{45}}+\frac{\mathrm{1}}{\mathrm{75}}+…\:=? \\ $$ Answered by EDWIN88 last updated on 02/Apr/21 $$\begin{array}{|c|c|c|c|c|}{}&\hline{\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{0}} }}&\hline{\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{1}} }}&\hline{\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }}&\hline{\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{3}} }}&\hline{…}&\hline{\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{n}} }}&\hline{\underset{\mathrm{n}=\mathrm{0}}…
Question Number 137374 by oooooooo last updated on 02/Apr/21 Answered by mr W last updated on 02/Apr/21 $${let}\:{OA}={OD}={R}={radius} \\ $$$$\left(\frac{{R}}{\:\sqrt{\mathrm{2}}}−\mathrm{4}\right)^{\mathrm{2}} +\left(\frac{{R}}{\:\sqrt{\mathrm{2}}}+\mathrm{3}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} \:\:\:…\left({i}\right) \\ $$$$\left(\frac{{R}}{\:\sqrt{\mathrm{2}}}−{x}\right)^{\mathrm{2}}…
Question Number 6302 by sanusihammed last updated on 22/Jun/16 Commented by Rasheed Soomro last updated on 22/Jun/16 $$\mathrm{3}^{{x}} +\mathrm{4}^{{x}} =\mathrm{25} \\ $$$$\mathrm{Since}\:\mathrm{3}^{\mathrm{x}} ,\:\mathrm{4}^{{x}} >\mathrm{0}\:\:\forall{x}\in\mathbb{Z} \\…
Question Number 71824 by mr W last updated on 20/Oct/19 $${solve}\: \\ $$$${x}^{{x}^{{x}^{\mathrm{2019}} } } =\mathrm{2019} \\ $$ Commented by mr W last updated on…