Question Number 2589 by Filup last updated on 23/Nov/15 $${f}_{{n}} =\frac{\mathrm{1}}{{n}}\left({n}+{f}_{{n}−\mathrm{1}} \right) \\ $$$${f}_{\mathrm{1}} =\mathrm{1} \\ $$$$ \\ $$$$\mathrm{Evaluate}: \\ $$$${S}=\underset{{i}=\mathrm{1}} {\overset{{m}} {\sum}}{f}_{{i}} \\ $$…
Question Number 2400 by Filup last updated on 19/Nov/15 $$\mathrm{According}\:\mathrm{to}\:\mathrm{Wikipedia}\:\mathrm{and}\:\mathrm{WolframAlpha}, \\ $$$$\mathrm{the}\:\mathrm{sign}\:\mathrm{function},\:\mathrm{sgn}\left({x}\right),\:\mathrm{is}\:\mathrm{defined}\:\mathrm{as}: \\ $$$$ \\ $$$$\mathrm{sgn}\left({x}\right)\equiv\frac{{x}}{\mid{x}\mid}=\frac{\mid{x}\mid}{{x}}\:\:\:\mathrm{for}\:{x}\neq\mathrm{0} \\ $$$$\mathrm{and}\:\mathrm{satisfies}: \\ $$$$\mathrm{sgn}\left({x}\right)=\sqrt{{x}}\sqrt{\frac{\mathrm{1}}{{x}}} \\ $$$$\boldsymbol{{but}} \\ $$$$\mathrm{sgn}\left(\mathrm{0}\right)=\mathrm{0} \\…
Question Number 67871 by rrebo5637@gmail.com last updated on 01/Sep/19 $$\mathrm{8}=\mathrm{4x} \\ $$$$\mathrm{x}=? \\ $$ Commented by Prithwish sen last updated on 01/Sep/19 $$\mathrm{2} \\ $$…
Question Number 2275 by Filup last updated on 13/Nov/15 $$\mathrm{Is}\:\mathrm{there}\:\mathrm{a}\:\mathrm{way}\:\mathrm{to}\:\mathrm{evaluate}\:\mathrm{the}\:\mathrm{following}: \\ $$$$ \\ $$$${S}=\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+…}}}} \\ $$ Answered by Rasheed Soomro last updated on 13/Nov/15 $${S}=\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+…}}}}…
Question Number 2181 by Filup last updated on 07/Nov/15 $$\mathrm{How}\:\mathrm{do}\:\mathrm{you}\:\mathrm{solve}\:\mathrm{the}\:\mathrm{following}\:\mathrm{series}? \\ $$$$\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{i}} \frac{\mathrm{1}}{\mathrm{2}\left({i}+\mathrm{1}\right)} \\ $$ Answered by prakash jain last updated on 08/Nov/15…
Question Number 133146 by shaker last updated on 19/Feb/21 Commented by Dwaipayan Shikari last updated on 19/Feb/21 $$\frac{\mathrm{1}−{a}^{{n}+\mathrm{1}} }{\mathrm{1}−{a}}\:\:\:\: \\ $$ Answered by mathmax by…
Question Number 67481 by Mohamed Amine Bouguezzoul last updated on 27/Aug/19 $${p}\:{is}\:{a}\:{prime}\:{number}\:{such}\:{that}\:\left(\mathrm{1}+{p}\right)^{{p}} \equiv\mathrm{2}\left[\mathrm{7}\right] \\ $$$${find}\:{all}\:{k}\:{such}\:{that}\:{p}\equiv{k}\left[\mathrm{42}\right] \\ $$ Commented by Rasheed.Sindhi last updated on 30/Aug/19 $$\boldsymbol{{Some}}\:{values}\:{of}\:{k}…
Question Number 1831 by Filup last updated on 10/Oct/15 $$\mathrm{Solve}: \\ $$$$\mathrm{1}+\left(\mathrm{1}+\mathrm{2}\right)+\left(\mathrm{1}+\mathrm{2}+\mathrm{3}\right)+…+\left(\mathrm{1}+\mathrm{2}+…+{n}\right) \\ $$ Answered by Rasheed Soomro last updated on 10/Oct/15 $$\mathrm{1}+\left(\mathrm{1}+\mathrm{2}\right)+\left(\mathrm{1}+\mathrm{2}+\mathrm{3}\right)+…+\left(\mathrm{1}+\mathrm{2}+…+{n}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}\right)\left(\mathrm{1}+\mathrm{1}\right)+\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}\right)\left(\mathrm{2}+\mathrm{1}\right)+…+\frac{\mathrm{1}}{\mathrm{2}}{n}\left({n}+\mathrm{1}\right)…
Question Number 66778 by Tony Lin last updated on 19/Aug/19 $$\frac{\mathrm{1}}{\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{2}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{5}}−\frac{\mathrm{2}}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{7}}+\frac{\mathrm{1}}{\mathrm{8}}−\frac{\mathrm{2}}{\mathrm{9}}+\frac{\mathrm{1}}{\mathrm{10}}+\frac{\mathrm{1}}{\mathrm{11}}−\frac{\mathrm{2}}{\mathrm{12}}+\centerdot\centerdot\centerdot= \\ $$ Commented by Prithwish sen last updated on 19/Aug/19 $$\left(\frac{\mathrm{1}}{\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}+……\right)−\left[\left(\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{3}}\right)+\left(\frac{\mathrm{1}}{\mathrm{6}}+\frac{\mathrm{2}}{\mathrm{6}}\right)+\left(\frac{\mathrm{1}}{\mathrm{9}}+\frac{\mathrm{2}}{\mathrm{9}}\right)+\left(\frac{\mathrm{1}}{\mathrm{12}}+\frac{\mathrm{2}}{\mathrm{12}}\right)+…\right] \\ $$$$=\underset{\mathrm{k}=\mathrm{1}} {\overset{\infty}…
Question Number 1181 by 22 last updated on 11/Jul/15 $$\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}=? \\ $$ Answered by Rasheed Ahmad last updated on 24/Jul/15 $${In}\:{certain}\:{cases}\:\sqrt{{a}+{b}\sqrt{{c}}\:}\:{can}\:{be} \\ $$$${simplified}\:{into}\:{p}+{q}\sqrt{{c}}\:{form}\left({not}\right. \\ $$$$\left.{in}\:{all}\:{cases}\right).\:{The}\:{procedure}\:{is}\:{as}…