Category: Arithmetic

Question Number 6158 by enigmeyou last updated on 16/Jun/16 $$\frac{-1}{1048576}=\frac{1}{8}\times {\left(\frac{-1}{2}\right)}^n$$$$findn?$$ Answered by Rasheed Soomro last updated on 16/Jun/16 $$\frac{-1}{1048576}=\frac{1}{8}\times {\left(\frac{-1}{2}\right)}^n$$$$\frac{−\mathrm{8}}{\mathrm{1048576}}=\left(\frac{−\mathrm{1}}{\mathrm{2}}\right)^{{n}}…

Question Number 137206 by JulioCesar last updated on 31/Mar/21 Answered by bemath last updated on 31/Mar/21 $$\mathrm{by}\mathrm{parts}\{\begin{array}{l}\mathrm{u}=\ln \left(\frac{\mathrm{x}-1}{\mathrm{x}+1}\right)\mathrm{du}=\frac{2}{(\mathrm{x}-1)(\mathrm{x}+1)}\mathrm{dx}\\ \mathrm{v}=\mathrm{x}\end{array}$$$$\mathrm{I}=\mathrm{x}\ln \left(\frac{\mathrm{x}-1}{\mathrm{x}+1}\right)-\int \frac{2\mathrm{x}}{(\mathrm{x}-1)(\mathrm{x}+1)}\mathrm{dx}$$$$\mathrm{I}=\mathrm{x}\ln \left(\frac{\mathrm{x}-1}{\mathrm{x}+1}\right)-[\int \frac{1}{\mathrm{x}-1}\mathrm{dx}+\int \frac{1}{\mathrm{x}+1}\mathrm{dx}]$$$$\mathrm{I}=\:\mathrm{x}\:\mathrm{ln}\:\left(\frac{\mathrm{x}−\mathrm{1}}{\mathrm{x}+\mathrm{1}}\right)−\mathrm{ln}\:\left(\mathrm{x}^{\mathrm{2}} −\mathrm{1}\right)\:+\:\mathrm{C}…

Question Number 137065 by mr W last updated on 29/Mar/21 Commented by mr W last updated on 29/Mar/21 $$whichnumberislarger?$$ Answered by mr W…

Question Number 137005 by greg_ed last updated on 28/Mar/21 $$\mathbf{Hi},\mathbf{guyz}!$$$$\mathbf{For}\mathbf{R}=\frac{1}{2}\times \frac{3}{4}\times \frac{5}{6}\times \dots \times \frac{223}{224}\mathbf{and}$$$$\mathbf{S}=\frac{2}{3}\times \frac{4}{5}\times \frac{6}{7}\times \dots \times \frac{224}{225}.$$$$\mathbf{Prove}\mathbf{that}:\mathbf{R}<\frac{1}{15}<\mathbf{S}.$$ Commented by greg_ed last updated on 01/May/21…

Question Number 5851 by sanusihammed last updated on 01/Jun/16 $$Solvesimultaneously$$$$$$$$2x+y-z=8\dots \dots \dots \left(i\right)$$$$x^2-y^2+2z^2=14\dots \dots \dots \left(ii\right)$$$$\mathrm{3}{x}^{\mathrm{3}} \:+\:\mathrm{4}{y}^{\mathrm{3}} \:+\:{z}^{\mathrm{3}} \:=\:\mathrm{195}\:\:\:\:………..\:\left({iii}\right) \…