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Category: Arithmetic

Given-U-n-2n-1-n-even-3n-3-n-odd-The-value-of-U-3-U-6-U-7-U-10-U-11-U-14-U-27-U-28-

Question Number 128907 by bramlexs22 last updated on 11/Jan/21 $$\:\mathrm{Given}\:\mathrm{U}_{\mathrm{n}} \:=\:\begin{cases}{\mathrm{2n}+\mathrm{1}\:;\:\mathrm{n}\:\mathrm{even}}\\{\mathrm{3n}+\mathrm{3}\:;\:\mathrm{n}\:\mathrm{odd}}\end{cases} \\ $$$$\mathrm{The}\:\mathrm{value}\:\mathrm{of}\:\mathrm{U}_{\mathrm{3}} +\mathrm{U}_{\mathrm{6}} +\mathrm{U}_{\mathrm{7}} +\mathrm{U}_{\mathrm{10}} +\mathrm{U}_{\mathrm{11}} + \\ $$$$\:\mathrm{U}_{\mathrm{14}} +…+\mathrm{U}_{\mathrm{27}} +\mathrm{U}_{\mathrm{28}} =? \\ $$…

coukd-you-prove-please-that-11-is-not-racional-number-

Question Number 128462 by Ari last updated on 07/Jan/21 $$\mathrm{coukd}\:\mathrm{you}\:\mathrm{prove}\:\mathrm{please}\:\mathrm{that}\:\sqrt{\mathrm{11}\:}\mathrm{is}\:\mathrm{not}\:\mathrm{racional}\:\mathrm{number} \\ $$ Answered by Dwaipayan Shikari last updated on 07/Jan/21 $$\sqrt{\mathrm{11}}=\mathrm{3}+\sqrt{\mathrm{11}}−\mathrm{3}=\mathrm{3}+\frac{\mathrm{2}}{\:\sqrt{\mathrm{11}}+\mathrm{3}}=\mathrm{3}+\frac{\mathrm{2}}{\mathrm{3}+\mathrm{3}+\frac{\mathrm{2}}{\mathrm{3}+\mathrm{3}+\frac{\mathrm{2}}{\mathrm{3}+\mathrm{3}+\frac{\mathrm{2}}{\mathrm{3}+\mathrm{3}…}}}} \\ $$$$=\mathrm{3}+\frac{\mathrm{2}}{\mathrm{6}+\frac{\mathrm{2}}{\mathrm{6}+\frac{\mathrm{2}}{\mathrm{6}+\frac{\mathrm{2}}{\mathrm{6}+\frac{\mathrm{2}}{\mathrm{6}+…}}}}}\:\:\:\:\: \\ $$$${It}\:{is}\:{an}\:{Infinte}\:{continued}\:{fraction}\:.{So}\:{it}\:{can}\:{never}\:{be}\:{Rational}…

Question-62570

Question Number 62570 by aliesam last updated on 23/Jun/19 Commented by mathmax by abdo last updated on 23/Jun/19 $${we}\:{have}\:{S}\left({x}\right)\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{sin}\left({nx}\right)}{\mathrm{2}^{{n}−\mathrm{1}} }\:=\mathrm{2}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{sin}\left({nx}\right)}{\mathrm{2}^{{n}} }\:=\mathrm{2}\:{Im}\left(\sum_{{n}=\mathrm{1}}…

show-that-n-N-12-divise-n-2-n-4-1-

Question Number 128012 by mathocean1 last updated on 03/Jan/21 $${show}\:{that}\:\forall\:{n}\:\in\:\mathbb{N},\:\mathrm{12}\:{divise}\:{n}^{\mathrm{2}} \left({n}^{\mathrm{4}} −\mathrm{1}\right) \\ $$ Answered by MJS_new last updated on 03/Jan/21 $${n}=\mathrm{2}{k}\:\Rightarrow\:\mathrm{2}\mid{n}\:\Rightarrow\:\mathrm{4}\mid{n}^{\mathrm{2}} \\ $$$${n}=\mathrm{2}{k}+\mathrm{1}\:\Rightarrow\:{n}^{\mathrm{4}} −\mathrm{1}=\mathrm{8}{k}\left({k}−\mathrm{1}\right)\left(\mathrm{2}{k}^{\mathrm{2}}…