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Question Number 9403 by alfat123 last updated on 05/Dec/16 Answered by ridwan balatif last updated on 05/Dec/16 $$\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{5x}\left(\mathrm{1}−\mathrm{x}\right)^{\mathrm{6}} \mathrm{dx}=\mathrm{u}.\mathrm{v}−\int\mathrm{vdu} \\ $$$$\mathrm{misal}:\:\mathrm{u}=\mathrm{5x}\rightarrow\mathrm{du}=\mathrm{5dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{dv}=\left(\mathrm{1}−\mathrm{x}\right)^{\mathrm{6}}…

Question Number 9396 by Shiva last updated on 04/Dec/16 $$\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}^{\mathrm{2}} }+\mathrm{4y}=\mathrm{tan}\:\mathrm{2x} \\ $$ Commented by 123456 last updated on 04/Dec/16 $${y}={y}_{{n}} +{y}_{{f}} \\ $$$${y}_{{n}}…

Question Number 140462 by ZiYangLee last updated on 08/May/21 Commented by mr W last updated on 08/May/21 $${T}_{{n}} ={ar}^{{n}−\mathrm{1}} \\ $$$$\frac{{T}_{{n}+\mathrm{4}} −{T}_{{n}+\mathrm{2}} }{{T}_{{n}+\mathrm{1}} }=\frac{{ar}^{{n}+\mathrm{3}} −{ar}^{{n}+\mathrm{1}}…

Question Number 9372 by geovane10math last updated on 03/Dec/16 $$\mathrm{I}\:\mathrm{was}\:\mathrm{able}\:\mathrm{to}\:\mathrm{discover}\:\mathrm{the}\:\mathrm{conditions}\:\mathrm{for} \\ $$$$\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{two}\:\mathrm{irrational}\:\mathrm{numbers}\:\mathrm{be}\:\mathrm{an} \\ $$$$\mathrm{integer}\:\mathrm{and}\:\mathrm{the}\:\mathrm{conditions}\:\mathrm{for}\:\mathrm{the}\:\mathrm{sum} \\ $$$$\mathrm{be}\:\mathrm{a}\:\mathrm{finite}\:\mathrm{decimal}. \\ $$$$\mathrm{But}\:\mathrm{I}\:\mathrm{can}\:\mathrm{not}\:\mathrm{do}\:\mathrm{the}\:\mathrm{same}\:\mathrm{for}\:\mathrm{periodic} \\ $$$$\mathrm{tithe}. \\ $$$$\mathrm{Someone}\:\mathrm{can}\:\mathrm{help}\:\mathrm{me},\:\mathrm{please}? \\ $$$$ \\…

Question Number 74863 by mrS last updated on 02/Dec/19 Commented by abdomathmax last updated on 03/Dec/19 $${we}\:{have}\:{S}=\sum_{{n}=\mathrm{1}} ^{\mathrm{45}} \:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:=\sum_{{p}=\mathrm{1}} ^{\left[\frac{\mathrm{45}}{\mathrm{2}}\right]} \:\:\frac{\mathrm{1}}{\left(\mathrm{2}{p}\right)^{\mathrm{2}} }\:+\sum_{{p}=\mathrm{0}} ^{\left[\frac{\mathrm{45}−\mathrm{1}}{\mathrm{2}}\right]} \:\frac{\mathrm{1}}{\left(\mathrm{2}{p}+\mathrm{1}\right)^{\mathrm{2}}…