Question Number 2655 by Yozzi last updated on 24/Nov/15 $${Prove}\:{by}\:{contradiction}\:{that}\:{there} \\ $$$${are}\:{no}\:{whole}\:{number}\:{solutions}\:\left({x},{y},{z}\right) \\ $$$${to}\:{the}\:{equation}\:{z}^{\mathrm{2}} ={x}^{\mathrm{2}} +{y}^{\mathrm{2}} \\ $$$${where}\:{both}\:{x}\:{and}\:{y}\:{are}\:{odd}. \\ $$ Answered by prakash jain last…
Question Number 2642 by Rasheed Soomro last updated on 24/Nov/15 $${n}\:{lines}\:{are}\:{drawn}\:{inside}\:{a}\:{circle}\:{in}\:{such}\:{a}\:{way}\:{that}\: \\ $$$${the}\:{circle}\:{has}\:{been}\:{divided}\:{in}\:{maximum}\:{number}\:{of} \\ $$$${parts}.\:{Determine}\:{this}\:{maximum}\:{number}. \\ $$ Commented by RasheedAhmad last updated on 24/Nov/15 $$\bullet{One}\:{line}\:{can}\:{divide}\:{the}\:{circle}…
Question Number 68161 by Learner-123 last updated on 06/Sep/19 $${In}\:{my}\:{textbook}\:{its}\:{written}: \\ $$$${In}\:{applying}\:{the}\:{nth}−{term}\:{test}\:{we}\: \\ $$$${can}\:{see}\:{that}: \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \:{diverges}\:{because}\: \\ $$$${lim}_{{n}\rightarrow\infty} \left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \:{does}\:{not}\:{exist}. \\ $$$${But}\:{then}\:{why}\:\underset{{n}=\mathrm{1}}…
Question Number 2619 by Rasheed Soomro last updated on 23/Nov/15 $${The}\:{sums}\:{of}\:{the}\:{first}\:\:{n}\:\:\:{terms}\:{of}\:{two}\:{AP}\:'{s}\:{are} \\ $$$${in}\:{the}\:{ratio}\:\:\mathrm{3}{n}+\mathrm{31}\::\:\:\mathrm{5}{n}−\mathrm{3}\:.\:{Show}\:{that}\:{their}\:\mathrm{9}^{{th}} \:{terms} \\ $$$${are}\:{equal}. \\ $$ Commented by Yozzi last updated on 24/Nov/15…
Question Number 133664 by shaker last updated on 23/Feb/21 Answered by liberty last updated on 23/Feb/21 $$\mathrm{partial}\:\mathrm{fraction} \\ $$$$\frac{\mathrm{1}}{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)}\:=\:\frac{\mathrm{A}}{\mathrm{x}+\mathrm{1}}+\frac{\mathrm{B}}{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{Cx}+\mathrm{D}}{\mathrm{x}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\Leftrightarrow\:\mathrm{1}=\left(\mathrm{x}+\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{2}}…
Question Number 68122 by TawaTawa last updated on 05/Sep/19 Commented by TawaTawa last updated on 05/Sep/19 $$\mathrm{Please}\:\mathrm{i}\:\mathrm{don}'\mathrm{t}\:\mathrm{understand}\:\mathrm{the}\:\mathrm{workings}\:\mathrm{here}.\:\:\mathrm{Help}\:\mathrm{me}\:\mathrm{explain}\:\mathrm{please} \\ $$ Answered by mind is power last…
Question Number 133653 by greg_ed last updated on 23/Feb/21 $$\boldsymbol{\mathrm{hi}},\:\boldsymbol{\mathrm{everybody}}\:! \\ $$$$\boldsymbol{\mathrm{how}}\:\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{prove}}\:\boldsymbol{\mathrm{that}}\:\boldsymbol{\pi}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{an}}\:\boldsymbol{\mathrm{irrational}}\:\boldsymbol{\mathrm{number}}\:??? \\ $$ Answered by Dwaipayan Shikari last updated on 23/Feb/21 $$\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{5}}−\frac{\mathrm{1}}{\mathrm{7}}+\frac{\mathrm{1}}{\mathrm{9}}−\frac{\mathrm{1}}{\mathrm{11}}+…=\frac{\pi}{\mathrm{4}} \\ $$$${As}\:{It}\:{is}\:{an}\:{Infinte}\:{series}\:{so}\:{it}\:{can}\:{never}\:{be}\:{rational}…
Question Number 2548 by Filup last updated on 22/Nov/15 $$\mathrm{For}\:\mathrm{a}\:\mathrm{function}\:{y}={f}\left({x}\right), \\ $$$$\mathrm{inflection}\:\mathrm{points}/\mathrm{stationary}\:\mathrm{points}\:\mathrm{are} \\ $$$$\mathrm{when}\:\:\frac{{df}}{{dx}}=\mathrm{0}. \\ $$$$ \\ $$$$\mathrm{For}\:\mathrm{a}\:\mathrm{function}\:{z}={f}\left({x},\:{y}\right),\:\mathrm{can}\:\mathrm{you}\:\mathrm{find} \\ $$$$\mathrm{these}\:\mathrm{points}\:\mathrm{through}\:\mathrm{a}\:\mathrm{similar}\:\mathrm{method}? \\ $$$$ \\ $$$$\mathrm{Is}\:\mathrm{it}\:\mathrm{something}\:\mathrm{like}\:\frac{\partial{f}}{\partial{x}}=\mathrm{0}\:\mathrm{and}\:\frac{\partial{f}}{\partial{y}}=\mathrm{0}? \\…
Question Number 2545 by Rasheed Soomro last updated on 22/Nov/15 $$\mathcal{P}{rove}\:{that}\:\mathrm{3}^{\mathrm{3}{n}} −\mathrm{26}{n}−\mathrm{1}\:{is}\:{divisible}\:{by}\:\mathrm{676}, \\ $$$${where}\:{n}\in\mathbb{Z}^{+} \\ $$ Commented by Yozzi last updated on 22/Nov/15 $${Use}\:{induction}. \\…
Question Number 2544 by Rasheed Soomro last updated on 22/Nov/15 $$\mathcal{C}{an}\:{you}\:\mathcal{G}{eneralize}\:{the}\:{following}? \\ $$$$\mathrm{1}+\mathrm{2}+\mathrm{3}+…+{n}=\frac{\mathrm{1}}{\mathrm{2}}\left({n}\right)\left({n}+\mathrm{1}\right) \\ $$$$\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +…+{n}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{6}}\left({n}\right)\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right. \\ $$$$\mathrm{1}^{\mathrm{3}} +\mathrm{2}^{\mathrm{3}} +\mathrm{3}^{\mathrm{3}} +…+{n}^{\mathrm{3}} =\left[\frac{\mathrm{1}}{\mathrm{2}}\left({n}\right)\left({n}+\mathrm{1}\right)\right]^{\mathrm{2}}…