Question Number 98280 by I want to learn more last updated on 12/Jun/20 $$\boldsymbol{\mathrm{Let}}\:\:\left\{\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{n}}} \right\}\:\:\boldsymbol{\mathrm{be}}\:\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{sequence}}\:\boldsymbol{\mathrm{such}}\:\boldsymbol{\mathrm{that}}\:\:\:\boldsymbol{\mathrm{a}}_{\mathrm{1}} \:=\:\:\mathrm{2}, \\ $$$$\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{n}}\:\:+\:\:\mathrm{1}} \:\:=\:\:\frac{\mathrm{3}\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{n}}} \:+\:\:\mathrm{4}}{\mathrm{2}\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{n}}} \:\:+\:\:\mathrm{3}},\:\:\:\:\:\boldsymbol{\mathrm{n}}\:\geqslant\:\mathrm{1},\:\:\:\:\:\boldsymbol{\mathrm{find}}\:\:\:\:\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{n}}} \\ $$ Answered by…
Question Number 163812 by mathlove last updated on 11/Jan/22 Answered by cortano1 last updated on 11/Jan/22 $$\:{p}\left({x}\right)=\:{x}^{\mathrm{3}} +\mathrm{1} \\ $$$$\:{p}\left(\mathrm{4}\right)=\:\mathrm{65} \\ $$ Commented by mathlove…
Question Number 98268 by mr W last updated on 16/Jun/20 $${let}\:{p}\left({x}\right)\:{be}\:{a}\:{polynomial}\:{function}\:{of} \\ $$$$\left({n}−\mathrm{1}\right)^{{th}} \:{degree}\:{and} \\ $$$${p}\left({k}\right)={k}\:{for}\:{k}=\mathrm{1},\mathrm{2},\mathrm{3},…,{n} \\ $$$${find}\:{p}\left(\mathrm{0}\right)\:{and}\:{p}\left({n}+\mathrm{1}\right). \\ $$$${example}:\:{n}=\mathrm{10} \\ $$ Commented by MJS…
Question Number 98267 by M±th+et+s last updated on 12/Jun/20 $$\forall\:{a},{b}>\mathrm{0}\:,\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{1} \\ $$$${prove}\:{that} \\ $$$$\left(\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}\right)\left(\frac{{b}}{{a}^{\mathrm{2}} +\mathrm{1}}+\frac{{a}}{{b}^{\mathrm{2}} +\mathrm{1}}\right)\geqslant\frac{\mathrm{8}}{\mathrm{3}} \\ $$ Answered by maths mind last…
Question Number 163782 by bekzodjumayev last updated on 10/Jan/22 Commented by bekzodjumayev last updated on 10/Jan/22 $${Please}\:{help} \\ $$ Answered by Ar Brandon last updated…
Question Number 163770 by mnjuly1970 last updated on 10/Jan/22 $$ \\ $$$$\:\:\:\:\:{solve}\::\:\:\:\:{x},{y}\:\in\:\mathbb{N}\: \\ $$$$\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{3}{x}\:+\:\mathrm{5}{y}\:=\:\mathrm{20}\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−−−−−−− \\ $$ Answered by MJS_new last updated…
Question Number 32682 by rahul 19 last updated on 31/Mar/18 $$\boldsymbol{{I}}{f}\:{x}_{\mathrm{1}} \:{and}\:{x}_{\mathrm{2}\:} \:{are}\:{roots}\:{of}\:{the}\:{equation} \\ $$$${acos}\:\mathrm{2}{x}+{bsin}\:{x}\:=\:{c}\:{and}\: \\ $$$$\mathrm{2}{sin}\:{x}_{\mathrm{1}} {sinx}_{\mathrm{2}} =\:{sin}\:{x}_{\mathrm{1}} +{sinx}_{\mathrm{2}} .\:\boldsymbol{{T}}{hen}\: \\ $$$${the}\:{value}\:{of}\:\:\frac{{b}}{{c}−{a}}\:{is}\:? \\ $$…
Question Number 32681 by rahul 19 last updated on 31/Mar/18 $$\boldsymbol{{T}}{otal}\:{no}.\:{of}\:{polynomials}\:{of}\:{the}\:{form} \\ $$$${x}^{\mathrm{3}} +{ax}^{\mathrm{2}} +{bx}+{c}\:\:{that}\:{are}\:{divisible}\:{by}\: \\ $$$${x}^{\mathrm{2}} +\mathrm{1},\:{where}\:{a},{b},{c}\in\mathrm{1},\mathrm{2},\mathrm{3},….,\mathrm{10}\:{is}\: \\ $$$$\left.\mathrm{1}\right)\:\mathrm{10} \\ $$$$\left.\mathrm{2}\right)\:\mathrm{15} \\ $$$$\left.\mathrm{3}\right)\:\mathrm{5} \\…
Question Number 98215 by I want to learn more last updated on 12/Jun/20 $$\boldsymbol{\mathrm{Find}}\:\boldsymbol{\mathrm{the}}\:\:\boldsymbol{\mathrm{nth}}\:\:\boldsymbol{\mathrm{term}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{sequence}}\:\:\left\{\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{n}}} \right\}\:\:\boldsymbol{\mathrm{such}}\:\boldsymbol{\mathrm{that}} \\ $$$$\:\:\:\:\frac{\boldsymbol{\mathrm{a}}_{\mathrm{1}} \:+\:\:\boldsymbol{\mathrm{a}}_{\mathrm{2}} \:+\:\:…\:\:+\:\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{n}}} }{\boldsymbol{\mathrm{n}}}\:\:\:=\:\:\boldsymbol{\mathrm{n}}\:\:+\:\:\frac{\mathrm{1}}{\boldsymbol{\mathrm{n}}}\:\:\left(\boldsymbol{\mathrm{n}}\:\:=\:\:\mathrm{1},\:\:\mathrm{2},\:\:\mathrm{3},\:\:…\right) \\ $$ Commented by Don08q…
Question Number 163747 by HongKing last updated on 10/Jan/22 $$\mathrm{a}^{\mathrm{2}} \left(\mathrm{a}\:-\:\mathrm{3b}\right)\:=\:\mathrm{b}^{\mathrm{2}} \left(\mathrm{b}\:-\:\mathrm{3a}\right)\:+\:\mathrm{27} \\ $$$$\mathrm{ab}\:=\:\mathrm{2} \\ $$$$\mathrm{find}\:\:\mathrm{a}^{\mathrm{3}} \:-\:\mathrm{b}^{\mathrm{3}} \:=\:? \\ $$ Answered by nurtani last updated…