Question Number 22503 by Tinkutara last updated on 19/Oct/17 $$\mathrm{If}\:\left({a}\:+\:{bx}\right)^{−\mathrm{2}} \:=\:\frac{\mathrm{1}}{\mathrm{4}}\:−\:\mathrm{3}{x}\:+\:…,\:\mathrm{then}\:\left({a},\:{b}\right)\:= \\ $$ Commented by mrW1 last updated on 19/Oct/17 $$\left(\mathrm{a}+\mathrm{bx}\right)^{−\mathrm{2}} =\mathrm{a}^{−\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{bx}}{\mathrm{a}}\right)^{−\mathrm{2}} =\mathrm{a}^{−\mathrm{2}} \left[\mathrm{1}−\mathrm{2}×\frac{\mathrm{bx}}{\mathrm{a}}+…\right]…
Question Number 153571 by mathdanisur last updated on 08/Sep/21 $$\mathrm{Find}\:\:\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}n}\centerdot\left(\frac{\pi^{\mathrm{2}} }{\mathrm{4}}\:-\:\mathrm{a}_{\boldsymbol{\mathrm{n}}} ^{\mathrm{2}} \right)\:=\:? \\ $$$$\mathrm{where}\:\:\mathrm{a}_{\boldsymbol{\mathrm{n}}} =\underset{\boldsymbol{\mathrm{k}}=\mathrm{1}} {\overset{\boldsymbol{\mathrm{n}}} {\sum}}\mathrm{arctan}\left(\frac{\mathrm{1}}{\mathrm{k}^{\mathrm{2}} -\mathrm{k}+\mathrm{1}}\right) \\ $$ Answered by mnjuly1970…
Question Number 22491 by Tinkutara last updated on 19/Oct/17 $$\mathrm{The}\:\mathrm{coefficient}\:\mathrm{of}\:{x}^{{r}} \:\mathrm{in}\:\mathrm{the}\:\mathrm{expansion}\:\mathrm{of} \\ $$$$\left(\mathrm{1}\:−\:\mathrm{2}{x}\right)^{−\mathrm{1}/\mathrm{2}} \:\mathrm{is} \\ $$$$\left(\mathrm{1}\right)\:\frac{\left(\mathrm{2}{r}\right)!}{\left({r}!\right)^{\mathrm{2}} } \\ $$$$\left(\mathrm{2}\right)\:\frac{\left(\mathrm{2}{r}\right)!}{\mathrm{2}^{{r}} \:\left({r}!\right)^{\mathrm{2}} } \\ $$$$\left(\mathrm{3}\right)\:\frac{\left(\mathrm{2}{r}\right)!}{\left({r}!\right)^{\mathrm{2}} \:\mathrm{2}^{\mathrm{2}{r}} }…
Question Number 88014 by ajfour last updated on 07/Apr/20 $${If}\:{there}\:{is}\:{no}\:{second}'{s}\:{hand}\:{on} \\ $$$${a}\:{clock}\:{and}\:{the}\:{minute}\:{and}\:{hour} \\ $$$${hand}\:{move}\:{in}\:{continuous}\:{fashion}, \\ $$$${then}\:{exactly}\:{at}\:{what}\:{time}\:{between} \\ $$$$\mathrm{02}:\mathrm{10}\:\:{and}\:\mathrm{02}:\mathrm{15}\:{does}\:{the}\:{position} \\ $$$${of}\:{the}\:{two}\:{hands}\:{exactly}\:{coincide}? \\ $$ Answered by mr…
Question Number 22474 by Tinkutara last updated on 19/Oct/17 $$\mathrm{Let}\:{R}\:=\:\left(\mathrm{5}\sqrt{\mathrm{5}}\:+\:\mathrm{11}\right)^{\mathrm{2}{n}+\mathrm{1}} \:\mathrm{and}\:{f}\:=\:{R}\:−\:\left[{R}\right], \\ $$$$\mathrm{then}\:\mathrm{prove}\:\mathrm{that}\:{Rf}\:=\:\mathrm{4}^{\mathrm{2}{n}+\mathrm{1}} . \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 22472 by Tinkutara last updated on 18/Oct/17 $$\mathrm{If}\:{x}^{{x}} \centerdot{y}^{{y}} \centerdot{z}^{{z}} \:=\:{x}^{{y}} \centerdot{y}^{{z}} \centerdot{z}^{{x}} \:=\:{x}^{{z}} \centerdot{y}^{{x}} \centerdot{z}^{{y}} \:\mathrm{such} \\ $$$$\mathrm{that}\:{x},\:{y}\:\mathrm{and}\:{z}\:\mathrm{are}\:\mathrm{positive}\:\mathrm{integers} \\ $$$$\mathrm{greater}\:\mathrm{than}\:\mathrm{1},\:\mathrm{then}\:\mathrm{which}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{following}\:\mathrm{cannot}\:\mathrm{be}\:\mathrm{true}\:\mathrm{for}\:\mathrm{any}\:\mathrm{of}\:\mathrm{the}…
Question Number 22468 by Tinkutara last updated on 18/Oct/17 $$\mathrm{If}\:{a}_{{r}} \:\mathrm{is}\:\mathrm{the}\:\mathrm{coefficient}\:\mathrm{of}\:{x}^{{r}} \:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{expansion}\:\left(\mathrm{1}\:+\:{x}\:+\:{x}^{\mathrm{2}} \right)^{{n}} ,\:\mathrm{then} \\ $$$${a}_{\mathrm{1}} \:−\:\mathrm{2}{a}_{\mathrm{2}} \:+\:\mathrm{3}{a}_{\mathrm{3}} \:−\:…….\:\mathrm{2}{na}_{\mathrm{2}{n}} \:= \\ $$ Terms…
Question Number 22463 by Tinkutara last updated on 18/Oct/17 $$\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:{x}: \\ $$$$\frac{\mathrm{1}}{\left[{x}\right]}\:+\:\frac{\mathrm{1}}{\left[\mathrm{2}{x}\right]}\:=\:\left({x}\right)\:+\:\frac{\mathrm{1}}{\mathrm{3}}, \\ $$$$\mathrm{where}\:\left[{x}\right]\:\mathrm{is}\:\mathrm{the}\:\mathrm{greatest}\:\mathrm{integer}\:\mathrm{less} \\ $$$$\mathrm{than}\:\mathrm{or}\:\mathrm{equal}\:\mathrm{to}\:{x}\:\mathrm{and}\:\left({x}\right)\:=\:{x}\:−\:\left[{x}\right], \\ $$$$\left[\mathrm{e}.\mathrm{g}.\:\left[\mathrm{3}.\mathrm{4}\right]\:=\:\mathrm{3}\:\mathrm{and}\:\left(\mathrm{3}.\mathrm{4}\right)\:=\:\mathrm{0}.\mathrm{4}\right]. \\ $$ Answered by ajfour last updated…
Question Number 87988 by M±th+et£s last updated on 07/Apr/20 $${solve}\:{the}\:{PDE} \\ $$$$\mathrm{1}−{Z}={px}+{py}−{q}\sqrt{{pq}} \\ $$$$\mathrm{2}−{Z}={px}+{qy}+{sin}\left({p}+{q}\right) \\ $$$$\mathrm{3}−{p}\left(\mathrm{1}+{q}^{\mathrm{2}} \right)={q}\left({Z}−{a}\right) \\ $$$$\mathrm{4}−{Z}={xyp}^{\mathrm{2}} \\ $$ Terms of Service Privacy…
Question Number 153513 by yeti123 last updated on 08/Sep/21 $$\left(\frac{{x}}{\mathrm{5}}\:+\:\frac{{y}}{\mathrm{3}}\right)\left(\frac{\mathrm{5}}{{x}}\:+\:\frac{\mathrm{3}}{{y}}\right)\:=\:\mathrm{139},\:\forall{x},{y}\:\in\:\mathbb{R}_{>\mathrm{0}} \\ $$$$\mathrm{find}\:\mathrm{maximum}\:\mathrm{and}\:\mathrm{minimum}\:\mathrm{of}\:\:\frac{{x}\:+\:{y}}{\:\sqrt{{xy}}} \\ $$ Answered by liberty last updated on 08/Sep/21 $$\Rightarrow\frac{{x}+{y}}{\:\sqrt{{xy}}}\:=\:\sqrt{\frac{{x}}{{y}}}+\sqrt{\frac{{y}}{{x}}}=\sqrt{\frac{\mathrm{5}\lambda}{\mathrm{3}}}+\sqrt{\frac{\mathrm{3}}{\mathrm{5}\lambda}} \\ $$$${from}\:{condition} \\…