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Author: Tinku Tara

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Question Number 60725 by naka3546 last updated on 25/May/19 $${x}\:\:=\:\:\underset{\mathrm{0}\leqslant{i}\leqslant{j}\leqslant\mathrm{2019}} {\sum}\:\left(\:_{\:\:\:\:{j}} ^{\mathrm{2019}} \:\right)\left(\:\:_{{i}} ^{{j}} \:\:\right)\: \\ $$ Commented by naka3546 last updated on 25/May/19 $${x}\:\:=\:\:?…

solve-for-x-a-a-x-a-a-x-2x-

Question Number 60723 by alphaprime last updated on 25/May/19 $${solve}\:{for}\:{x}\: \\ $$$$\sqrt{{a}−\sqrt{{a}+{x}}}\:+\:\sqrt{{a}+\sqrt{{a}−{x}}}\:=\:\mathrm{2}{x} \\ $$ Commented by MJS last updated on 25/May/19 $$\mathrm{what}'\mathrm{s}\:\mathrm{the}\:\mathrm{source}\:\mathrm{of}\:\mathrm{this}\:\mathrm{problem}? \\ $$ Commented…

Question-60717

Question Number 60717 by peter frank last updated on 24/May/19 Commented by Prithwish sen last updated on 25/May/19 $$=\begin{vmatrix}{\frac{\partial\mathrm{x}}{\partial\mathrm{r}}\:\:\:\:\:\frac{\partial\mathrm{x}}{\partial\theta}\:\:\:\:\:\frac{\partial\mathrm{x}}{\partial\emptyset}}\\{\frac{\partial\mathrm{y}}{\partial\mathrm{r}}\:\:\:\:\frac{\partial\mathrm{y}}{\partial\theta}\:\:\:\:\:\:\frac{\partial\mathrm{y}}{\partial\phi}}\end{vmatrix} \\ $$$$\:\:\:\:\begin{vmatrix}{\frac{\partial\mathrm{z}}{\partial\mathrm{r}}\:\:\:\:\:\frac{\partial\mathrm{z}}{\partial\theta}\:\:\:\:\:\:\frac{\partial\mathrm{z}}{\partial\phi}}\end{vmatrix} \\ $$$$=\mathrm{sin}\theta\mathrm{cos}\phi\left(\mathrm{r}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \theta\:\mathrm{cos}\phi\:\right)+\mathrm{sin}\theta\mathrm{sin}\phi\left(\mathrm{r}^{\mathrm{2}}…

prove-that-2x-4-2-3-4-3x-5-3-4-5-4x-6-4-5-6-100x-102-100-101-102-103-102-

Question Number 191790 by mathlove last updated on 30/Apr/23 $${prove}\:{that} \\ $$$$\frac{\mathrm{2}{x}−\mathrm{4}}{\mathrm{2}\centerdot\mathrm{3}\centerdot\mathrm{4}}+\frac{\mathrm{3}{x}−\mathrm{5}}{\mathrm{3}\centerdot\mathrm{4}\centerdot\mathrm{5}}+\frac{\mathrm{4}{x}−\mathrm{6}}{\mathrm{4}\centerdot\mathrm{5}\centerdot\mathrm{6}}+…..+\frac{\mathrm{100}{x}−\mathrm{102}}{\mathrm{100}\centerdot\mathrm{101}\centerdot\mathrm{102}}=\frac{\mathrm{103}}{\mathrm{102}} \\ $$$$ \\ $$ Commented by BaliramKumar last updated on 01/May/23 $$\mathrm{if}\:\mathrm{x}=\mathrm{1}\:\mathrm{then}\:\mathrm{all}\:\mathrm{term}\:\left(−\mathrm{ve}\right)\:\mathrm{but}\:\frac{\mathrm{103}}{\mathrm{102}}\:\mathrm{is}\:\left(+\mathrm{ve}\right)\: \\…

Question-126253

Question Number 126253 by mohammad17 last updated on 18/Dec/20 Answered by physicstutes last updated on 18/Dec/20 $$\mathrm{Exactly} \\ $$$$\:{a}\:+\:{ar}\:+\:{ar}^{\mathrm{2}} \:+…+{ar}^{{n}−\mathrm{1}} \:=\:\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{ar}^{{n}−\mathrm{1}} \:=\:\frac{{a}\left(\mathrm{1}−{r}^{{n}} \right)}{\mathrm{1}−{r}}…

Question-60716

Question Number 60716 by peter frank last updated on 24/May/19 Commented by maxmathsup by imad last updated on 25/May/19 $${let}\:{A}\left({x}\right)\:=\left(\frac{{x}}{{x}+\mathrm{1}}\right)^{{x}} \:\Rightarrow{A}\left({x}\right)\:=\left(\mathrm{1}−\frac{{x}+\mathrm{1}}{{x}}\right)^{{x}} \:={e}^{{x}\:{ln}\left(\mathrm{1}−\frac{\mathrm{1}}{{x}+\mathrm{1}}\right)} \\ $$$${but}\:{we}\:{have}\:{for}\:{u}\:\in{V}\left(\mathrm{0}\right)\:{ln}\left(\mathrm{1}−{u}\right)\:\sim−{u}\:\Rightarrow{ln}\left(\mathrm{1}−\frac{\mathrm{1}}{{x}+\mathrm{1}}\right)\sim−\frac{\mathrm{1}}{{x}+\mathrm{1}}\:{for}\:{x}\in{V}\left(+\infty\right) \\…