Question Number 3385 by Filup last updated on 12/Dec/15 $$\mathrm{let}: \\ $$$$\:\:\:\:\:\:\:\:{S}=\underset{{i}={a}} {\overset{{n}} {\sum}}{x}_{{i}} \\ $$$$\mathrm{prove}\:\mathrm{or}\:\mathrm{disprove}\:\mathrm{that}: \\ $$$$\:\:\:\:\:\:\frac{{d}}{{dx}}\left(\underset{{i}={a}} {\overset{{n}} {\sum}}{x}_{{i}} \right)=\underset{{i}={a}} {\overset{{n}} {\sum}}\left(\frac{{d}}{{dx}}\left\{{x}_{{i}} \right\}\right) \\…
Question Number 68825 by peter frank last updated on 15/Sep/19 Answered by mind is power last updated on 16/Sep/19 $${dx}\left(\mathrm{1}+{e}^{\frac{{x}}{{y}}} \right)+{e}^{\frac{{x}}{{y}}} \left(\mathrm{1}−\frac{{x}}{{y}}\right){dy} \\ $$$${p}=\mathrm{1}+{e}^{\frac{{x}}{{y}}} \\…
Question Number 3275 by Filup last updated on 09/Dec/15 $$\mathrm{Prove}\:\mathrm{to}\:\mathrm{me}\:\mathrm{that}\:\mathrm{2}\:\mathrm{is}\:\mathrm{the}\:{only}\:\mathrm{even} \\ $$$$\mathrm{prime}\:\mathrm{number} \\ $$ Commented by Filup last updated on 09/Dec/15 $$\mathrm{For}\:\mathrm{prime}\:{P}\in\mathbb{P} \\ $$$$\mathrm{if}\:\:{P}=\frac{{x}}{{y}} \\…
Question Number 68783 by peter frank last updated on 15/Sep/19 Commented by peter frank last updated on 22/Sep/19 $${thank}\:{you}\: \\ $$ Commented by ajfour last…
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Question Number 68775 by rajesh4661kumar@gmail.com last updated on 15/Sep/19 Answered by $@ty@m123 last updated on 15/Sep/19 $${Let}\:{a}\:{be}\:\:{the}\:{first}\:{term}\:{and}\:{d}\:{be}\:{the} \\ $$$${common}\:{difference}\:{of}\:{AP}. \\ $$$${a}+\left({p}−\mathrm{1}\right){d}={A}\:…\left(\mathrm{1}\right) \\ $$$${a}+\left({q}−\mathrm{1}\right){d}={AR}\:\:….\left(\mathrm{2}\right) \\ $$$${a}+\left({r}−\mathrm{1}\right){d}={AR}^{\mathrm{2}}…
Question Number 3204 by Filup last updated on 07/Dec/15 $$\underset{{i}=\mathrm{1}} {\overset{{n}} {\prod}}{i}!=\mathrm{1}×\left(\mathrm{1}×\mathrm{2}\right)×…×\left(\mathrm{1}×\mathrm{2}×…×{n}\right) \\ $$$$=\mathrm{1}^{{n}} \mathrm{2}^{{n}−\mathrm{1}} \mathrm{3}^{{n}−\mathrm{2}} …\left({n}−\mathrm{1}\right)^{\mathrm{2}} {n} \\ $$$$\therefore\:\underset{{i}=\mathrm{1}} {\overset{{n}} {\prod}}{i}!=\underset{{i}=\mathrm{1}} {\overset{{n}} {\prod}}\left({n}−{i}+\mathrm{1}\right)^{{i}} \\…
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Question Number 3198 by Filup last updated on 07/Dec/15 $$\underset{{i}={a}} {\overset{{n}} {\prod}}{i}=?\:\:\:\:\:\mathrm{for}\:\mid{i}\mid<\mathrm{1},\:\:{a}\in\mathbb{R} \\ $$ Commented by Filup last updated on 07/Dec/15 $$=\frac{\mathrm{1}}{{a}}×\frac{\mathrm{1}}{{a}+\mathrm{1}}×…×\frac{\mathrm{1}}{{n}} \\ $$$$=\frac{\mathrm{1}}{\left(\frac{{n}!}{\left({a}−\mathrm{1}\right)!}\right)} \\…