Question Number 150404 by mathdanisur last updated on 12/Aug/21 $$\Omega\:=\underset{\:\mathrm{0}} {\overset{\:\infty} {\int}}\:\frac{\mathrm{log}\left(\mathrm{x}\right)}{\left(\mathrm{1}\:+\:\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\:\mathrm{dx}\:=\:? \\ $$ Answered by Ar Brandon last updated on 12/Aug/21 $$\Omega\left({a}\right)=\int_{\mathrm{0}}…
Question Number 19332 by Tinkutara last updated on 09/Aug/17 $$\mathrm{Let}\:{S}_{{n}} \:=\:{n}^{\mathrm{2}} \:+\:\mathrm{20}{n}\:+\:\mathrm{12},\:{n}\:\mathrm{a}\:\mathrm{positive} \\ $$$$\mathrm{integer}.\:\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{all}\:\mathrm{possible} \\ $$$$\mathrm{values}\:\mathrm{of}\:{n}\:\mathrm{for}\:\mathrm{which}\:{S}_{{n}} \:\mathrm{is}\:\mathrm{a}\:\mathrm{perfect} \\ $$$$\mathrm{square}? \\ $$ Commented by mrW1 last…
Question Number 150402 by mathdanisur last updated on 12/Aug/21 $$\mathrm{For}\:\:\mathrm{m}\geqslant\mathrm{1} \\ $$$$\boldsymbol{\Omega}\:=\underset{\:\mathrm{0}} {\overset{\:\infty} {\int}}\:\frac{\mathrm{x}\:\mathrm{ln}^{\boldsymbol{\mathrm{m}}} \:\left(\mathrm{x}\right)}{\mathrm{e}^{\boldsymbol{\mathrm{x}}} \:−\:\mathrm{1}}\:=\:\mathrm{2}\:\mathrm{ln}^{\boldsymbol{\mathrm{m}}} \:\boldsymbol{\zeta}\left(\mathrm{3}\right) \\ $$ Commented by yadiirama last updated on…
Question Number 19329 by Tinkutara last updated on 09/Aug/17 $$\mathrm{Solve}\:\mathrm{the}\:\mathrm{equation}\:{y}^{\mathrm{3}} \:=\:{x}^{\mathrm{3}} \:+\:\mathrm{8}{x}^{\mathrm{2}} \:−\:\mathrm{6}{x}\:+\:\mathrm{8} \\ $$$$\mathrm{for}\:\mathrm{positive}\:\mathrm{integers}\:{x}\:\mathrm{and}\:{y}. \\ $$ Commented by Tinkutara last updated on 12/Aug/17 $$\mathrm{RMO}\:\mathrm{2000}…
Question Number 19313 by Tinkutara last updated on 09/Aug/17 $$\mathrm{Prove}\:\mathrm{that}\:\mid{z}_{\mathrm{1}} \:\pm\:{z}_{\mathrm{2}} \mid^{\mathrm{2}} \:=\:\mid{z}_{\mathrm{2}} \mid^{\mathrm{2}} \:+\:\mid{z}_{\mathrm{1}} \mid^{\mathrm{2}} \:\pm \\ $$$$\mathrm{2Re}\left({z}_{\mathrm{1}} \bar {{z}}_{\mathrm{2}} \right)\:=\:\mid{z}_{\mathrm{1}} \mid^{\mathrm{2}} \:+\:\mid{z}_{\mathrm{2}} \mid^{\mathrm{2}}…
Question Number 19312 by Tinkutara last updated on 09/Aug/17 $$\mathrm{Product}\:\mathrm{of}\:{n},\:{n}^{\mathrm{th}} \:\mathrm{roots}\:\mathrm{of}\:\mathrm{unity} \\ $$$$=\:\mathrm{1}.\alpha.\alpha^{\mathrm{2}} .\alpha^{\mathrm{3}} \:…..\:\alpha^{{n}−\mathrm{1}} \:=\:\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \\ $$$$\mathrm{Why}?\:\mathrm{How}\:\mathrm{to}\:\mathrm{get}\:\mathrm{RHS}? \\ $$ Answered by ajfour last updated…
Question Number 150374 by mathdanisur last updated on 11/Aug/21 $$\underset{\boldsymbol{\mathrm{k}}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\mathrm{2}^{\boldsymbol{\mathrm{k}}} \:+\:\mathrm{3}^{\boldsymbol{\mathrm{k}}} }{\mathrm{5}^{\boldsymbol{\mathrm{k}}} }\:\:=\:? \\ $$ Answered by eman_64 last updated on 11/Aug/21 $$\:\:\:=\:\sum_{\mathrm{k}=\mathrm{0}}…
Question Number 84835 by Power last updated on 16/Mar/20 Commented by abdomathmax last updated on 16/Mar/20 $${S}=\sum_{{k}=\mathrm{1}} ^{\mathrm{32}} \:\frac{{k}}{{k}+\mathrm{1}}\:=\sum_{{k}=\mathrm{1}} ^{\mathrm{32}} \:\frac{{k}+\mathrm{1}−\mathrm{1}}{{k}+\mathrm{1}} \\ $$$$=\sum_{{k}=\mathrm{1}} ^{\mathrm{32}} \left(\mathrm{1}\right)−\sum_{{k}=\mathrm{1}}…
Question Number 84820 by Power last updated on 16/Mar/20 Answered by MJS last updated on 16/Mar/20 $$\mathrm{super}\:\mathrm{easy} \\ $$$$\sqrt{\mathrm{2}{x}−\mathrm{5}}={t}\:\Leftrightarrow\:{x}=\frac{\mathrm{1}}{\mathrm{2}}{t}^{\mathrm{2}} +\frac{\mathrm{5}}{\mathrm{2}}\wedge{t}\geqslant\mathrm{0} \\ $$$$\sqrt{\frac{\mathrm{1}}{\mathrm{2}}{t}^{\mathrm{2}} +\frac{\mathrm{5}}{\mathrm{2}}−\mathrm{2}+{t}}+\sqrt{\frac{\mathrm{1}}{\mathrm{2}}{t}^{\mathrm{2}} +\frac{\mathrm{5}}{\mathrm{2}}+\mathrm{2}+\mathrm{3}{t}}=\mathrm{7}\sqrt{\mathrm{2}} \\…
Question Number 150350 by mathdanisur last updated on 11/Aug/21 Terms of Service Privacy Policy Contact: info@tinkutara.com