Question Number 194960 by Erico last updated on 20/Jul/23 $$\mathrm{Soit}\:{x}>\mathrm{1}.\:\mathrm{On}\:\mathrm{d}\acute {\mathrm{e}finie}\:\mathrm{la}\:\mathrm{suite}\:\left(\mathrm{p}_{\mathrm{n}} \right)\:\mathrm{par}\: \\ $$$$\mathrm{p}_{\mathrm{1}} ={x}\:\:\mathrm{et}\:\forall\mathrm{n}\in\mathrm{IN}^{\ast} \:\:\:\:\:\mathrm{p}_{\mathrm{n}+\mathrm{1}} =\mathrm{2p}_{\mathrm{n}} ^{\mathrm{2}} −\mathrm{1} \\ $$$$\mathrm{Montrer}\:\mathrm{que}\:\underset{\mathrm{n}\rightarrow+\infty} {\mathrm{lim}}\:\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\prod}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{p}_{\mathrm{k}} }\right)=\sqrt{\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}}…
Question Number 194638 by Erico last updated on 12/Jul/23 $$\mathrm{Prove}\:\mathrm{that}\:\forall{n}\in\mathrm{IN}^{\ast} \:\:\:\:\: \\ $$$$\:\:\:\underset{{k}=\mathrm{1}} {\overset{\mathrm{2}^{{n}} −\mathrm{1}} {\sum}}\:\frac{\mathrm{1}}{{sin}^{\mathrm{2}} \left(\frac{{k}\pi}{\mathrm{2}^{{n}+\mathrm{1}} }\right)}=\:\frac{\mathrm{2}^{\mathrm{2}{n}+\mathrm{1}} −\mathrm{2}}{\mathrm{3}} \\ $$$$\mathrm{Give}\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:{n}\:\:\:\underset{{k}=\mathrm{1}} {\overset{\mathrm{2}^{{n}} −\mathrm{1}} {\sum}}\:\frac{\mathrm{1}}{{sin}^{\mathrm{4}} \left(\frac{{k}\pi}{\mathrm{2}^{{n}+\mathrm{1}}…
Question Number 193864 by pascal889 last updated on 21/Jun/23 Answered by MM42 last updated on 21/Jun/23 $$\frac{\frac{{n}!}{\left({n}−\mathrm{5}\right)!}}{\frac{{n}!}{\mathrm{4}!\left({n}−\mathrm{4}\right)!}}=\mathrm{144}\Rightarrow{n}−\mathrm{4}=\mathrm{6}\Rightarrow{n}=\mathrm{10}\checkmark \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 193368 by MATHEMATICSAM last updated on 11/Jun/23 $$\mathrm{If}\:\mathrm{log}_{{a}} {y}\:=\:\frac{\mathrm{1}}{\mathrm{3}}\:\mathrm{and}\:\mathrm{log}_{\mathrm{8}} {a}\:=\:{x}\:+\:\mathrm{1}\:\mathrm{then}\:\mathrm{show} \\ $$$$\mathrm{that}\:{y}\:=\:\mathrm{2}^{{x}\:+\:\mathrm{1}} \\ $$ Answered by aba last updated on 11/Jun/23 $$\mathrm{log}_{\mathrm{8}} \mathrm{a}=\mathrm{x}+\mathrm{1}\:\Rightarrow\:\mathrm{ln}\left(\mathrm{a}\right)=\mathrm{3}\left(\mathrm{x}+\mathrm{1}\right)\mathrm{ln}\left(\mathrm{2}\right)\:\Rightarrow\:\mathrm{ln}\left(\mathrm{a}\right)=\mathrm{3ln}\left(\mathrm{2}^{\mathrm{x}+\mathrm{1}}…
Question Number 65405 by KanhAshish last updated on 29/Jul/19 Commented by KanhAshish last updated on 29/Jul/19 $$\mathrm{someone}\:\mathrm{please}\:\mathrm{send}\:\mathrm{explanation}. \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 130899 by greg_ed last updated on 30/Jan/21 $$\mathrm{There}\:\mathrm{are}\:\mathrm{20}\:\mathrm{persons}\:\mathrm{at}\:\mathrm{a}\:\mathrm{party}. \\ $$$$\mathrm{Maria}\:\mathrm{dansed}\:\mathrm{with}\:\mathrm{7}\:\mathrm{boys}, \\ $$$$\mathrm{Stacy}\:\mathrm{dansed}\:\mathrm{with}\:\mathrm{8}\:\mathrm{boys}, \\ $$$$\mathrm{Monia}\:\mathrm{dansed}\:\mathrm{with}\:\mathrm{9}\:\mathrm{boys}\:\mathrm{and} \\ $$$$\mathrm{Eve}\:\mathrm{dansed}\:\mathrm{with}\:\mathrm{all}\:\mathrm{the}\:\mathrm{boys}\:\mathrm{at}\:\mathrm{the}\:\mathrm{party}. \\ $$$$\boldsymbol{\mathrm{How}}\:\boldsymbol{\mathrm{many}}\:\boldsymbol{\mathrm{girls}}\:\boldsymbol{\mathrm{are}}\:\boldsymbol{\mathrm{there}}\:\boldsymbol{\mathrm{at}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{party}}\:? \\ $$ Commented by mr…
Question Number 130689 by liberty last updated on 28/Jan/21 $$ \\ $$$$\mathrm{Mr}.\:\mathrm{Rahmat}\:\mathrm{decided}\:\mathrm{to}\:\mathrm{create}\:\mathrm{a}\:\mathrm{password}\: \\ $$$$\mathrm{with}\:\mathrm{the}\:\mathrm{form}\:\mathrm{numbers}\:\mathrm{and}\:\mathrm{letterse} \\ $$$$\mathrm{intermittntly}\:\:\mathrm{intermittent}\:\left(\:\mathrm{can}\:\mathrm{also}\:\mathrm{be}\right. \\ $$$$\left.\mathrm{letters}\:\mathrm{and}\:\mathrm{numbers}\:\mathrm{intermittently}\right)\:\mathrm{but}\:\mathrm{no} \\ $$$$\mathrm{nmbers}\:\mathrm{and}\:\mathrm{letters}\:\mathrm{are}\:\mathrm{the}\:\mathrm{same}.\:\mathrm{It}\: \\ $$$$\mathrm{chooses}\:\mathrm{to}\:\mathrm{use}\:\mathrm{numbers}\:\mathrm{on}\:\mathrm{the}\:\mathrm{set}\:\:\left\{\mathrm{2},\:\mathrm{5},\:\mathrm{8},\:\mathrm{9}\right\}\: \\ $$$$\mathrm{and}\:\mathrm{selects}\:\mathrm{the}\:\mathrm{letters}\:\mathrm{on}\:\mathrm{the}\:\mathrm{set}\:\left\{\:\mathrm{A},\:\mathrm{X},\:\mathrm{Y},\:\mathrm{W},\mathrm{Z}\right\}\: \\…
Question Number 130685 by EDWIN88 last updated on 28/Jan/21 $${How}\:{many}\:{password}\:{containing} \\ $$$$\mathrm{6}\:{characters}\:{of}\:{the}\:{letters}\:{in}\:{the}\: \\ $$$${word}\:''\:{Move}\:{Now}\:''\:? \\ $$ Answered by liberty last updated on 28/Jan/21 $$\mathrm{case}\left(\mathrm{1}\right)\:\underset{−} {\mathrm{O}}\underset{−}…
Question Number 130634 by EDWIN88 last updated on 27/Jan/21 Answered by liberty last updated on 27/Jan/21 $$\mathrm{coefficient}\:\mathrm{of}\:\mathrm{x}^{\mathrm{8}} \:\mathrm{from}\:\left(\mathrm{1}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}!}+\frac{\mathrm{x}^{\mathrm{4}} }{\mathrm{4}!}+\frac{\mathrm{x}^{\mathrm{6}} }{\mathrm{6}!}+\frac{\mathrm{x}^{\mathrm{8}} }{\mathrm{8}!}\right)^{\mathrm{2}} \\ $$$$\mathrm{is}\:\frac{\mathrm{2}}{\mathrm{8}!}+\frac{\mathrm{2}}{\mathrm{2}!\mathrm{6}!}+\frac{\mathrm{1}}{\mathrm{4}!\mathrm{4}!}=\frac{\mathrm{1}}{\mathrm{315}} \\…
Question Number 130602 by EDWIN88 last updated on 27/Jan/21 Answered by mr W last updated on 27/Jan/21 $$={number}\:{of}\:\mathrm{4}\:{digit}\:{numbers}\:{which} \\ $$$${are}\:{divisible}\:{by}\:\mathrm{3}: \\ $$$$\mathrm{1002},\:\mathrm{1005},\:…,\:\mathrm{9999} \\ $$$$\Rightarrow\frac{\mathrm{9999}−\mathrm{1002}}{\mathrm{3}}+\mathrm{1}=\mathrm{3000} \\…