Question Number 219135 by Rojarani last updated on 20/Apr/25 Answered by Frix last updated on 20/Apr/25 $${a}^{\frac{\mathrm{1}}{\mathrm{3}}} +{b}^{\frac{\mathrm{1}}{\mathrm{3}}} ={c}^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$${a}+{b}+\mathrm{3}{a}^{\frac{\mathrm{1}}{\mathrm{3}}} {b}^{\frac{\mathrm{1}}{\mathrm{3}}} \underset{={c}^{\frac{\mathrm{1}}{\mathrm{3}}} } {\underbrace{\left({a}^{\frac{\mathrm{1}}{\mathrm{3}}}…
Question Number 219181 by hardmath last updated on 20/Apr/25 Answered by devdutt last updated on 20/Apr/25 $$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{k}^{\mathrm{2}} }{\mathrm{2}{k}^{\mathrm{2}} −\mathrm{2}{nk}+{n}^{\mathrm{2}} }\:=\:\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{{k}^{\mathrm{2}} }{{k}^{\mathrm{2}}…
Question Number 219112 by hardmath last updated on 19/Apr/25 $$\mathrm{Prove}\:\mathrm{it}: \\ $$$$\mathrm{In}\:\mathrm{triangle}\:\mathrm{ABC},\:\mathrm{AB}=\mathrm{c},\:\mathrm{BC}=\mathrm{b},\:\mathrm{AC}=\mathrm{a} \\ $$$$\mathrm{ab}^{\mathrm{2}} \mathrm{c}\:+\:\mathrm{abc}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} \mathrm{bc}\:\geqslant\:\mathrm{tan}\:\frac{\mathrm{A}}{\mathrm{2}}\:\frac{\mathrm{2S}^{\mathrm{3}} }{\left(\mathrm{a}+\mathrm{b}\right)^{\mathrm{2}} −\mathrm{c}^{\mathrm{2}} } \\ $$ Terms of Service…
Question Number 219113 by hardmath last updated on 19/Apr/25 $$\frac{\mathrm{a}\:+\:\mathrm{3b}}{\mathrm{a}\:+\:\mathrm{b}−\mathrm{1}}\:+\:\frac{\mathrm{a}\:+\:\mathrm{3b}−\mathrm{1}}{\mathrm{a}\:+\:\mathrm{b}−\mathrm{3}}\:=\:\mathrm{4}\:\:\Rightarrow\:\:\mathrm{a}\:+\:\mathrm{b}\:=\:? \\ $$ Commented by Rasheed.Sindhi last updated on 19/Apr/25 $$\mathrm{a}+\mathrm{b}\:\mathrm{is}\:\mathrm{not}\:\mathrm{unique},\: \\ $$ Answered by Rasheed.Sindhi…
Question Number 219110 by universe last updated on 19/Apr/25 $${find}\:{the}\:{nth}\:{term}\:{of}\:{x}_{{n}+\mathrm{1}} \:=\:{x}_{{n}} \left(\mathrm{2}−{x}_{{n}} \right) \\ $$$${in}\:{x}_{\mathrm{1}} \\ $$ Answered by vnm last updated on 22/Apr/25 $${x}_{{n}+\mathrm{1}}…
Question Number 219093 by hardmath last updated on 19/Apr/25 Commented by hardmath last updated on 19/Apr/25 $$\mathrm{ABC}\:=\:\bigtriangleup \\ $$$$\mathrm{B}\:-\:\mathrm{acute}\:\mathrm{angle} \\ $$$$\angle\mathrm{B}\:=\:\mathrm{2}\:\centerdot\:\angle\mathrm{C} \\ $$$$\mathrm{AB}\:=\:\mathrm{10} \\ $$$$\mathrm{BC}\:=\:\mathrm{22}…
Question Number 219025 by universe last updated on 18/Apr/25 Answered by vnm last updated on 19/Apr/25 $${the}\:{generating}\:{function}\:{of}\: \\ $$$${this}\:{sequence}\:{is} \\ $$$${f}\left({x}\right)=\frac{{x}−\mathrm{2ln}\left(\mathrm{1}−{x}\right)+\frac{\mathrm{1}}{\mathrm{1}−{x}}−\mathrm{1}}{\left(\mathrm{2}−{x}\right)^{\mathrm{2}} } \\ $$ Commented…
Question Number 219003 by hardmath last updated on 17/Apr/25 Answered by maths2 last updated on 18/Apr/25 $${S}_{{p}} =\Sigma\Sigma\underset{{k}=\mathrm{0}} {\sum}\frac{{k}^{\mathrm{2}} }{\left({n}−{k}\right)^{\mathrm{2}} +{k}^{\mathrm{2}} }=\Sigma\Sigma\underset{{k}=\mathrm{0}} {\sum}\frac{\left({n}−{k}\right)^{\mathrm{2}} }{\left({n}−{k}\right)^{\mathrm{2}} +{k}^{\mathrm{2}}…
Question Number 218997 by hardmath last updated on 17/Apr/25 Answered by MrGaster last updated on 19/Apr/25 $$\mathrm{4}!\underset{\alpha+\beta+\gamma+\delta=\mathrm{4}} {\sum}\frac{\left[{x}\right]^{\alpha} \left[{x}/\mathrm{2}\right]^{\beta} \left[{x}/\mathrm{4}\right]^{\gamma} \left[{x}/\mathrm{8}\right]^{\delta} }{\alpha!\beta!\gamma!\delta!}=\left(\left[{x}\right]+\left[\frac{{x}}{\mathrm{2}}\right]+\left[\frac{{x}}{\mathrm{4}}\right]+\left[\frac{{x}}{\mathrm{8}}\right]\right)^{\mathrm{4}} \\ $$$$\Rightarrow\left(\left[{x}\right]+\left[\frac{{x}}{\mathrm{2}}\right]+\left[\frac{{x}}{\mathrm{4}}\right]\left[\frac{{x}}{\mathrm{8}}\right]\right)^{\mathrm{4}} \equiv\mathrm{1}\:\left(\mathrm{mod}\:\mathrm{5}\right)\:\left(^{\ast}…
Question Number 218970 by Nicholas666 last updated on 17/Apr/25 $$ \\ $$$$\:\:\:\:\:\:\:\:\mathrm{2},\mathrm{12},\mathrm{18},\mathrm{48},\mathrm{50},….. \\ $$$$ \\ $$ Answered by Frix last updated on 17/Apr/25 $$\mathrm{The}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{42}\:\mathrm{as}\:\mathrm{always}. \\…