Question Number 197951 by universe last updated on 05/Oct/23 $$\:\underset{{n}=\mathrm{1}\:} {\overset{\infty} {\sum}}\frac{{n}}{{n}^{\mathrm{4}} +{n}^{\mathrm{2}} +\mathrm{1}}\:−\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{{n}^{\mathrm{2}} }{{n}^{\mathrm{8}} +{n}^{\mathrm{4}} +\mathrm{1}}\:=\:? \\ $$ Terms of Service Privacy…
Question Number 197935 by Frix last updated on 04/Oct/23 $$\mathrm{Show}\:\mathrm{that} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\left({n}!\right)^{\mathrm{2}} }{\left(\mathrm{2}{n}\right)!}\:=\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{2}\pi\sqrt{\mathrm{3}}}{\mathrm{27}} \\ $$ Answered by witcher3 last updated on 05/Oct/23 $$\underset{\mathrm{n}\geqslant\mathrm{1}}…
Question Number 197922 by cherokeesay last updated on 04/Oct/23 Commented by Frix last updated on 04/Oct/23 $$\mathrm{Question}\:\mathrm{197876} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 197880 by universe last updated on 02/Oct/23 $$\mathrm{find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{infinite}\:\mathrm{series} \\ $$$$\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{1}} }\centerdot\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\centerdot\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{4}} }\left(\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} \right)\:+\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }\centerdot\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{6}} }\left(\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +…+\mathrm{7}^{\mathrm{2}} \right)+…
Question Number 197876 by cortano12 last updated on 02/Oct/23 Commented by Frix last updated on 02/Oct/23 $$\mathrm{You}\:\mathrm{can}\:\mathrm{only}\:\mathrm{approximate} \\ $$$${x}\approx.\mathrm{707349599414} \\ $$$${y}\approx.\mathrm{654856853342} \\ $$$$\mathrm{No}\:\mathrm{other}\:\mathrm{solution}\:\in\mathbb{C} \\ $$…
Question Number 197895 by tri26112004 last updated on 02/Oct/23 $${Solve}\:{the}\:{equation}: \\ $$$${x}^{\mathrm{4}} \:−\:{x}^{\mathrm{3}} \:−\:\mathrm{4}{x}^{\mathrm{2}} \:+\:\mathrm{3}{x}\:+\:\mathrm{2}\:=\:\mathrm{0} \\ $$ Answered by Frix last updated on 02/Oct/23 $$\left({x}−\mathrm{2}\right)\left({x}^{\mathrm{3}}…
Question Number 197860 by cortano12 last updated on 01/Oct/23 $$\:\:\:\:\frac{\mathrm{2}^{\mathrm{17}} +\mathrm{2}^{\mathrm{16}} +\mathrm{2}^{\mathrm{15}} +\ldots+\mathrm{1}}{\mathrm{2}^{\mathrm{8}} +\mathrm{2}^{\mathrm{7}} +\mathrm{2}^{\mathrm{6}} +\ldots+\mathrm{1}}\:=\:?\: \\ $$ Answered by mr W last updated on…
Question Number 197843 by a.lgnaoui last updated on 30/Sep/23 $$\boldsymbol{\mathrm{Exercice}}\:\:\mathrm{2} \\ $$ Commented by a.lgnaoui last updated on 30/Sep/23 Commented by witcher3 last updated on…
Question Number 197838 by hardmath last updated on 30/Sep/23 Answered by MM42 last updated on 30/Sep/23 $$\frac{\mathrm{2}^{\mathrm{49}} }{\mathrm{2}^{\mathrm{49}} +\mathrm{1}}+\frac{\mathrm{2}^{\mathrm{48}} }{\mathrm{2}^{\mathrm{48}} +\mathrm{1}}+…+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{48}} +\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{49}} +\mathrm{1}} \\ $$$$=\mathrm{1}+\mathrm{1}+…+\mathrm{1}=\:\mathrm{49}\:\checkmark…
Question Number 197784 by cortano12 last updated on 28/Sep/23 $$\:\sqrt[{\mathrm{3}}]{\mathrm{x}−\mathrm{2}+\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{x}−\mathrm{3}}\:\left(\mathrm{1}+\sqrt[{\mathrm{3}}]{\mathrm{x}−\mathrm{3}}\right)}\:+\:\sqrt[{\mathrm{3}}]{\mathrm{x}+\mathrm{5}+\mathrm{6}\sqrt[{\mathrm{3}}]{\mathrm{x}−\mathrm{3}}\left(\mathrm{1}+\mathrm{2}\sqrt[{\mathrm{3}}]{\mathrm{x}−\mathrm{3}}\:\right)}\:=\:\mathrm{5} \\ $$ Answered by Frix last updated on 28/Sep/23 $$\mathrm{Let}\:{x}={t}^{\mathrm{3}} +\mathrm{3} \\ $$$$\sqrt[{\mathrm{3}}]{\left({t}+\mathrm{1}\right)^{\mathrm{3}} }+\sqrt[{\mathrm{3}}]{{t}^{\mathrm{3}} +\mathrm{12}{t}^{\mathrm{2}}…