# Category: Operation Research

Question Number 130827 by bemath last updated on 29/Jan/21 $$\left[\:\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}\:} \mathrm{D}^{\mathrm{2}} +\left(\mathrm{x}+\mathrm{1}\right)\mathrm{D}+\mathrm{1}\:\right]\mathrm{y}\:=\:\mathrm{4cos}\:\left(\mathrm{ln}\left(\:\mathrm{x}+\mathrm{1}\right)\right) \\$$ Answered by EDWIN88 last updated on 29/Jan/21 $${let}\:\mathrm{ln}\:\left({x}+\mathrm{1}\right)={t}\:\Rightarrow{x}+\mathrm{1}\:=\:{e}^{{t}} \\$$$$\begin{cases}{\frac{{dy}}{{dx}}\:=\:\frac{{dy}}{{dt}}×\frac{{dt}}{{dx}}\:=\:\frac{\mathrm{1}}{{x}+\mathrm{1}}.\frac{{dy}}{{dt}}}\\{\frac{{d}^{\mathrm{2}} {y}}{{dx}}\:=\:\frac{{d}}{{dx}}\:\left[\:\frac{\mathrm{1}}{{x}+\mathrm{1}}\:\frac{{dy}}{{dt}}\:\right]=\:\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}}… ## Solve-134-x-1-16-x-768- Question Number 129131 by Boucatchou last updated on 13/Jan/21$${Solve}\:\:\mathrm{134}^{{x}+\mathrm{1}} =\mathrm{16}^{{x}} −\mathrm{768} \\ $$Commented by Dwaipayan Shikari last updated on 13/Jan/21$${No}\:{solution}\:{in}\:\:{x}\in\mathbb{R} \\ $$… ## Question-129102 Question Number 129102 by AbdulAzizabir last updated on 12/Jan/21 Answered by Olaf last updated on 12/Jan/21$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}^{\mathrm{2}} \:=\:\left\{\mathrm{1},\:\mathrm{5},\:\mathrm{14},\:\mathrm{30},\:\mathrm{55},…\right\}\:=\:\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}} \\ \mathrm{S}\:=\:\frac{\mathrm{2}}{\mathrm{3}}\left(\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{20}}+\frac{\mathrm{1}}{\mathrm{56}}−\frac{\mathrm{1}}{\mathrm{120}}+\frac{\mathrm{1}}{\mathrm{220}}−…\right) \\ \mathrm{S}\:=\:\frac{\mathrm{1}}{\mathrm{6}}\left(\frac{\mathrm{1}}{\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{14}}−\frac{\mathrm{1}}{\mathrm{30}}+\frac{\mathrm{1}}{\mathrm{55}}−…\right) \\…

Question Number 63298 by Rio Michael last updated on 02/Jul/19 $${A}\:{particle}\:{P},\:{moves}\:{on}\:{the}\:{curve}\:{with}\:{polar}\:{equation}\:\: \\$$$${r}\:=\:{e}^{{k}\theta} \:,\:{where}\:\left({r},\theta\right)\:{are}\:{polar}\:{coordinates}\:{referred}\:{to}\:{a}\:{fixed} \\$$$${pole}\:{and}\:{k}\:{is}\:{a}\:{positive}\:{constant}.\:{Given}\:{that}\:{the}\:{radial}\:{velocity} \\$$$${of}\:{P}\:{is}\:\frac{{k}}{{r}}\:\:{show}\:{that}\:{the}\:{transverse}\:{acceleration}\:{of}\:{th}\:{particle} \\$$$${is}\:{zero}. \\$$$$\\$$ Commented…

## A-n-1-5-n-1-5-n-A-n-

Question Number 124041 by pticantor last updated on 30/Nov/20 $$\\$$$$\\$$$$\\$$$$\:\:\:\:\:\boldsymbol{{A}}_{{n}} =\left(\mathrm{1}+\sqrt{\mathrm{5}}\right)^{\boldsymbol{{n}}} −\left(\mathrm{1}−\sqrt{\mathrm{5}}\right)^{\boldsymbol{{n}}} \\$$$$\\$$$$\:\:\:\boldsymbol{{A}}_{\boldsymbol{{n}}} =???? \\$$…

## W-n-k-1-2n-k-n-2-k-2-montrer-que-W-n-converge-et-calculer-la-valeur-de-W-n-

Question Number 123826 by pticantor last updated on 28/Nov/20 $$\\$$$$\boldsymbol{{W}}_{\boldsymbol{{n}}} =\underset{\boldsymbol{{k}}=\mathrm{1}} {\overset{\mathrm{2}\boldsymbol{{n}}} {\sum}}\frac{\boldsymbol{{k}}}{\boldsymbol{{n}}^{\mathrm{2}} +\boldsymbol{{k}}^{\mathrm{2}} } \\$$$$\boldsymbol{{montrer}}\:\boldsymbol{{que}}\:\boldsymbol{{W}}_{\boldsymbol{{n}}} \:\boldsymbol{{converge}} \\$$$$\boldsymbol{{et}}\:\boldsymbol{{calculer}}\:\boldsymbol{{la}}\:\boldsymbol{{valeur}}\:\boldsymbol{{de}}\:\boldsymbol{{W}}_{\boldsymbol{{n}}} \\$$ Commented…