Question Number 82019 by M±th+et£s last updated on 17/Feb/20 Answered by mind is power last updated on 18/Feb/20 $$\mathrm{2}{F}\mathrm{1}\left({k}+\mathrm{1},{k}+\mathrm{1};{k}+\mathrm{2};\mathrm{1}\right)=\left({k}+\mathrm{1}\right)\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{k}} \left(\mathrm{1}−{x}\right)^{−{k}−\mathrm{1}} {dx} \\ $$$${not}\:{defind}\:{sir}\:{tchek}\:{this}\:{Quation}…
Question Number 16450 by ajfour last updated on 22/Jun/17 $${Ten}\:{balls}\:{were}\:{manufactured}, \\ $$$$\:{nine}\:{of}\:{them}\:{have}\:{the}\:{same} \\ $$$${mass},\:{while}\:{just}\:{one}\:{of}\:{them} \\ $$$${has}\:{a}\:{slightly}\:{higher}\:{or}\:{slightly} \\ $$$${lower}\:{mass}.\:{Given}\:{is}\:{just}\:{a}\:{beam} \\ $$$${balance}\:{and}\:{no}\:{weights}.\:{comparing} \\ $$$${the}\:{masses}\:{of}\:{balls}\:{only},\:{with}\:{the} \\ $$$${help}\:{of}\:{the}\:{balance}\:,\:{and}\:{in}\:{just} \\…
Question Number 147493 by henderson last updated on 21/Jul/21 $$\boldsymbol{\mathrm{hi}},\:\boldsymbol{\mathrm{dears}}\:\boldsymbol{\mathrm{masters}}\:! \\ $$$$\boldsymbol{\mathrm{A}}\:=\:\left\{\boldsymbol{{au}}+\boldsymbol{{bv}},\:\left(\boldsymbol{{a}},\:\boldsymbol{{b}},\:\boldsymbol{{u}},\:\boldsymbol{{v}}\right)\:\in\:\mathbb{Z}^{\mathrm{4}} \right\}\:\boldsymbol{\mathrm{with}}\:\boldsymbol{{a}}\:\neq\:\boldsymbol{{b}}. \\ $$$$\mathrm{1}.\:\boldsymbol{\mathrm{prove}}\:\boldsymbol{\mathrm{that}}\:\boldsymbol{\mathrm{A}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{ideal}}\:\boldsymbol{\mathrm{of}}\:\:\mathbb{Z}. \\ $$$$\mathrm{2}.\:\boldsymbol{\mathrm{let}}\:\boldsymbol{\lambda}\mathbb{Z}\:=\:\left\{\boldsymbol{\lambda{n}}\:,\:\boldsymbol{{n}}\:\in\:\mathbb{Z}\right\}.\: \\ $$$$\boldsymbol{\mathrm{prove}}\:\boldsymbol{\mathrm{that}}\:\boldsymbol{\mathrm{A}}\:\boldsymbol{\mathrm{has}}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{smaller}}\:\boldsymbol{\mathrm{element}}\:\boldsymbol{\lambda}\:\boldsymbol{\mathrm{strictly}}\: \\ $$$$\boldsymbol{\mathrm{positive}}\:\boldsymbol{\mathrm{such}}\:\boldsymbol{\mathrm{that}}\:\boldsymbol{\mathrm{A}}\:=\:\boldsymbol{\lambda}\mathbb{Z}. \\ $$$$\mathrm{3}.\:\boldsymbol{\mathrm{prove}}\:\boldsymbol{\mathrm{that}}\:\boldsymbol{\lambda}\:=\:\boldsymbol{\mathrm{gcd}}\left(\boldsymbol{{a}},\boldsymbol{{b}}\right). \\ $$…
Question Number 81911 by jagoll last updated on 16/Feb/20 Commented by jagoll last updated on 16/Feb/20 $${a}_{{n}+\mathrm{1}} \:=\:\frac{\mathrm{3}{a}_{{n}} +\mathrm{1}}{\mathrm{2}}\: \\ $$$${a}_{{n}+\mathrm{1}} \:+{A}\:=\:{A}+\frac{\mathrm{3}{a}_{{n}} +\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{3}{a}_{{n}} +\mathrm{2}{A}+\mathrm{1}}{\mathrm{2}} \\…
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Question Number 147334 by Tawa11 last updated on 19/Jul/21 $$\mathrm{If}\:\:\:\:\mathrm{log}_{\mathrm{6}} \mathrm{30}\:\:\:=\:\:\:\mathrm{a},\:\:\:\:\:\:\:\:\:\mathrm{log}_{\mathrm{15}} \mathrm{24}\:\:\:=\:\:\:\mathrm{b},\:\:\:\:\:\:\:\:\:\:\:\mathrm{evaluate}:\:\:\:\mathrm{log}_{\mathrm{12}} \mathrm{60} \\ $$ Commented by otchereabdullai@gmail.com last updated on 20/Jul/21 $$\mathrm{nice}! \\ $$…
Question Number 147294 by ArielVyny last updated on 19/Jul/21 $$\Sigma\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{4}} }=? \\ $$ Answered by mathmax by abdo last updated on 19/Jul/21 $$\sum_{\mathrm{n}=\mathrm{1}} ^{\infty}…
Question Number 147252 by mathlove last updated on 19/Jul/21 Answered by Olaf_Thorendsen last updated on 19/Jul/21 $${l}\:=\:\left(\frac{\mathrm{1}}{\mathrm{5}}+\left(\frac{\mathrm{1}}{\mathrm{5}}+\left(\frac{\mathrm{1}}{\mathrm{5}}+…\right)^{\mathrm{2}} \right)^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$${l}\:=\:\left(\frac{\mathrm{1}}{\mathrm{5}}+{l}\right)^{\mathrm{2}} \\ $$$${l}^{\mathrm{2}} −\frac{\mathrm{3}}{\mathrm{5}}{l}+\frac{\mathrm{1}}{\mathrm{25}}\:=\:\mathrm{0} \\…
Question Number 81446 by john santu last updated on 13/Feb/20 $$\mathrm{given}\:\mathrm{a}_{\mathrm{1}} \:=\:\mathrm{2}\:,\:\mathrm{a}_{\mathrm{2}} \:=\:\mathrm{3}\:\mathrm{and} \\ $$$$\mathrm{a}_{\mathrm{n}+\mathrm{2}} \:=\:\mathrm{a}_{\mathrm{n}+\mathrm{1}} \:+\:\frac{\mathrm{a}_{\mathrm{n}} }{\mathrm{2}} \\ $$$$\mathrm{find}\:\mathrm{a}_{\mathrm{n}} \:=? \\ $$ Commented by…