Question Number 8013 by Yozzia last updated on 28/Sep/16 $${Prove}\:{that},\:\forall{m}\in\mathbb{N}, \\ $$$$\underset{{r}=\mathrm{1}} {\overset{{m}} {\sum}}\begin{pmatrix}{\mathrm{2}{m}}\\{\mathrm{2}{r}−\mathrm{1}}\end{pmatrix}\:\left(\mathrm{2}{r}−\mathrm{1}\right)^{\mathrm{2}{m}−\mathrm{1}} =\underset{{r}=\mathrm{1}} {\overset{{m}} {\sum}}\begin{pmatrix}{\mathrm{2}{m}}\\{\mathrm{2}{r}}\end{pmatrix}\:\left(\mathrm{2}{r}\right)^{\mathrm{2}{m}−\mathrm{1}} . \\ $$ Commented by FilupSmith last updated…
Question Number 139076 by bramlexs22 last updated on 22/Apr/21 $$\lceil\:\mathrm{3x}\:+\:\mathrm{5}\lfloor\:\mathrm{x}\:\rfloor\:\rceil\:=\:\mathrm{9} \\ $$ Answered by mathmax by abdo last updated on 22/Apr/21 $$\Rightarrow\left[\mathrm{3x}\right]+\mathrm{5}\left[\mathrm{x}\right]=\mathrm{9}\:\:\:\mathrm{let}\:\left[\mathrm{x}\right]=\mathrm{n}\:\Rightarrow\mathrm{n}\leqslant\mathrm{x}<\mathrm{n}+\mathrm{1}\:\Rightarrow\mathrm{3n}\leqslant\mathrm{3x}<\mathrm{3n}+\mathrm{3} \\ $$$$\mathrm{if}\:\mathrm{3n}\leqslant\mathrm{3x}<\mathrm{3n}+\mathrm{1}\:\Rightarrow\left[\mathrm{3x}\right]=\mathrm{3n}\:\:\mathrm{and}\:\mathrm{e}\Rightarrow\mathrm{3n}+\mathrm{5n}=\mathrm{9}\:\Rightarrow\mathrm{8n}=\mathrm{9}\:\:\mathrm{impossible} \\…
Question Number 8003 by Yozzia last updated on 27/Sep/16 $$\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\left(\underset{{r}=\mathrm{0}} {\overset{{m}−\mathrm{1}} {\sum}}\begin{pmatrix}{\mathrm{2}{m}}\\{\mathrm{2}{r}+\mathrm{1}}\end{pmatrix}\:\left(−\mathrm{1}\right)^{{m}−{r}} \right)^{\mathrm{2}} +\left(\underset{{r}=\mathrm{0}} {\overset{{m}} {\sum}}\begin{pmatrix}{\mathrm{2}{m}}\\{\mathrm{2}{r}}\end{pmatrix}\:\left(−\mathrm{1}\right)^{{m}−{r}} \right)^{\mathrm{2}} }\right)=? \\ $$ Commented by prakash…
Question Number 7999 by FilupSmith last updated on 27/Sep/16 $${S}=\frac{\mathrm{1}}{{n}}+\frac{\mathrm{1}}{\mathrm{2}{n}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{4}{n}^{\mathrm{4}} }+\frac{\mathrm{1}}{\mathrm{8}{n}^{\mathrm{8}} }+… \\ $$$${S}=\frac{\mathrm{1}}{{n}}+\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}^{{i}} {n}^{\mathrm{2}^{{i}} } } \\ $$$$\mathrm{Solvable}? \\ $$ Answered…
Question Number 139057 by EnterUsername last updated on 21/Apr/21 $$\mathrm{The}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{region}\:\mathrm{in}\:\mathrm{the}\:\mathrm{complex}\:\mathrm{plane}\:\mathrm{satisfying} \\ $$$$\mathrm{the}\:\mathrm{inequality}\:\mathrm{log}_{\mathrm{cos}\left(\frac{\pi}{\mathrm{6}}\right)} \left[\frac{\mid\mathrm{z}−\mathrm{2}\mid+\mathrm{5}}{\mathrm{4}\mid\mathrm{z}−\mathrm{2}\mid−\mathrm{4}}\right]<\mathrm{2}\:\mathrm{is}\:? \\ $$ Answered by MJS_new last updated on 22/Apr/21 $$\mid{z}−\mathrm{2}\mid={x}\geqslant\mathrm{0} \\ $$$$\frac{\mathrm{ln}\:\frac{{x}+\mathrm{5}}{\mathrm{4}\left({x}−\mathrm{1}\right)}}{\mathrm{ln}\:\mathrm{cos}\:\frac{\pi}{\mathrm{6}}}<\mathrm{2}\:\Leftrightarrow\:\mathrm{ln}\:\frac{{x}+\mathrm{5}}{{x}−\mathrm{1}}\:>\mathrm{ln}\:\mathrm{3}…
Question Number 139052 by EnterUsername last updated on 21/Apr/21 $$\left(\mathrm{1}+\mathrm{z}\right)^{\mathrm{n}} =\left(\mathrm{1}−\mathrm{z}\right)^{\mathrm{n}} \\ $$$${where}\:{z}\:{is}\:{a}\:{complex}\:{number} \\ $$ Answered by mathmax by abdo last updated on 21/Apr/21 $$\mathrm{z}=−\mathrm{1}\:\mathrm{is}\:\mathrm{not}\:\mathrm{solution}\:\:\mathrm{let}\:\mathrm{z}\neq−\mathrm{1}…
Question Number 139051 by mathsuji last updated on 21/Apr/21 $${a};{b};{c}\in\mathbb{N} \\ $$$$\left({a}+{b}\right)\left({a}+\mathrm{2}{b}\right)\left({a}+\mathrm{3}{b}\right)=\mathrm{105}^{\boldsymbol{{c}}} \\ $$ Commented by Rasheed.Sindhi last updated on 22/Apr/21 $$\left({a}+{b}\right)\left({a}+\mathrm{2}{b}\right)\left({a}+\mathrm{3}{b}\right)=\mathrm{105}^{\boldsymbol{{c}}} \\ $$$$\left({a}+{b}\right)\left({a}+\mathrm{2}{b}\right)\left({a}+\mathrm{3}{b}\right)=\mathrm{3}^{{c}} .\mathrm{5}^{{c}}…
Question Number 139055 by EnterUsername last updated on 21/Apr/21 $$\mathrm{Let}\:\mathrm{a}\:\mathrm{and}\:\mathrm{b}\:\mathrm{be}\:\mathrm{complex}\:\mathrm{numbers}\:\mathrm{representing}\:\mathrm{the}\:\mathrm{points} \\ $$$$\mathrm{A}\:\mathrm{and}\:\mathrm{B}\:\mathrm{respectively}\:\mathrm{in}\:\mathrm{the}\:\mathrm{complex}\:\mathrm{plane}. \\ $$$$\mathrm{If}\:\frac{\mathrm{a}}{\mathrm{b}}+\frac{\mathrm{b}}{\mathrm{a}}=\mathrm{1}\:\mathrm{and}\:\mathrm{O}\:\mathrm{is}\:\mathrm{the}\:\mathrm{origin}.\:\mathrm{Then}\:\Delta\mathrm{OAB}\:\mathrm{is}\:? \\ $$ Answered by MJS_new last updated on 22/Apr/21 $$\frac{{a}}{{b}}+\frac{{b}}{{a}}=\mathrm{1}\:\Rightarrow\:{b}={a}\left(\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right) \\…
Question Number 7969 by tawakalitu last updated on 26/Sep/16 $${During}\:{one}\:{year}\:{in}\:{a}\:{school},\:\frac{\mathrm{5}}{\mathrm{8}}\:{of}\:{the}\:{students} \\ $$$${had}\:{measiles}.\:\frac{\mathrm{1}}{\mathrm{2}}\:{had}\:{chickenpox},\:{and}\:\frac{\mathrm{1}}{\mathrm{8}}\:{had} \\ $$$${Neither}.\:{What}\:{fraction}\:{of}\:{the}\:{school}\:{had}\:{both} \\ $$$${measiles}\:{and}\:{chickenpox}. \\ $$ Answered by Rasheed Soomro last updated on…
Question Number 73468 by mathocean1 last updated on 12/Nov/19 $$\mathrm{soit}\:\mathrm{le}\:\mathrm{systeme}\:\mathrm{suivant} \\ $$$$\begin{cases}{\mathrm{2s}+\mathrm{4c}+\mathrm{3t}=\mathrm{700}}\\{\mathrm{3s}+\mathrm{2c}+\mathrm{2t}=\mathrm{500}}\end{cases} \\ $$$$\:\:\mathrm{8s}+\mathrm{7c}+\mathrm{8t}=…?… \\ $$$$\mathrm{comment}\:\mathrm{determiner}\:\mathrm{le}\:\mathrm{resultat}\:…?…\: \\ $$$$\mathrm{de}\:\mathrm{la}\:\mathrm{3}^{\mathrm{e}} \mathrm{equation}\:? \\ $$ Answered by MJS last…