Question Number 17187 by virus last updated on 02/Jul/17 Answered by Tinkutara last updated on 02/Jul/17 $$\frac{{a}_{\mathrm{1}} \:+\:{a}_{\mathrm{2}} \:+\:…\:+\:{a}_{{p}} }{{a}_{\mathrm{1}} \:+\:{a}_{\mathrm{2}} \:+\:…\:+\:{a}_{{q}} }\:=\:\frac{\frac{{p}}{\mathrm{2}}\left[\mathrm{2}{a}_{\mathrm{1}} \:+\:\left({p}\:−\:\mathrm{1}\right){d}\right]}{\frac{{q}}{\mathrm{2}}\left[\mathrm{2}{a}_{\mathrm{1}} \:+\:\left({q}\:−\:\mathrm{1}\right){d}\right]}…
Question Number 148219 by iloveisrael last updated on 26/Jul/21 $$\mathrm{The}\:\mathrm{expansion}\:\mathrm{of}\:\left(\mathrm{1}+\mathrm{px}+\mathrm{qx}^{\mathrm{2}} \right)^{\mathrm{8}} \: \\ $$$$=\:\mathrm{1}+\mathrm{8x}+\mathrm{52x}^{\mathrm{2}} +\mathrm{kx}^{\mathrm{3}} +… \\ $$$$\mathrm{What}\:\mathrm{are}\:\mathrm{the}\:\mathrm{values}\:\mathrm{of}\:\mathrm{p}\:,\mathrm{q}\:\mathrm{and}\:\mathrm{k} \\ $$ Answered by liberty last updated…
Question Number 17102 by tawa tawa last updated on 30/Jun/17 $$\mathrm{compute}:\:\:\:\underset{\mathrm{k}\:=\:\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\mathrm{2k}\:+\:\mathrm{1}}{\mathrm{2}^{\mathrm{2}\left(\mathrm{k}\:+\:\mathrm{1}\right)} } \\ $$ Commented by prakash jain last updated on 01/Jul/17 $$\mathrm{S}=\:\underset{\mathrm{k}\:=\:\mathrm{0}}…
Question Number 17075 by tawa tawa last updated on 30/Jun/17 $$\mathrm{Given}\:\mathrm{that}:\:\:\mathrm{log}\left(\frac{\mathrm{x}}{\mathrm{y}\:−\:\mathrm{z}}\right)\:=\:\mathrm{log}\left(\frac{\mathrm{y}}{\mathrm{z}\:−\:\mathrm{x}}\right)\:=\:\mathrm{log}\left(\frac{\mathrm{z}}{\mathrm{x}\:−\:\mathrm{y}}\right) \\ $$$$\mathrm{Show}\:\mathrm{that}\::\:\:\:\mathrm{x}^{\mathrm{x}} \:×\:\mathrm{y}^{\mathrm{y}} \:×\:\mathrm{z}^{\mathrm{z}} \:=\:\mathrm{1} \\ $$ Commented by RasheedSoomro last updated on 30/Jun/17…
Question Number 82520 by peter frank last updated on 22/Feb/20 Commented by mathmax by abdo last updated on 22/Feb/20 $${let}\:{S}_{{n}} =\sum_{{k}=\mathrm{0}} ^{{n}} \:\mathrm{2}^{{k}} \:{cos}\left(\mathrm{2}{k}\theta\right)\:\Rightarrow{S}_{{n}} ={Re}\left(\sum_{{k}=\mathrm{0}}…
Question Number 148048 by puissant last updated on 25/Jul/21 $$\mathrm{2}^{\mathrm{x}} −\mathrm{3}^{\left(\frac{\mathrm{x}}{\mathrm{2}}\right)} =\mathrm{1} \\ $$ Answered by mindispower last updated on 25/Jul/21 $$\mathrm{2}^{{x}} −\left(\sqrt{\mathrm{3}}\right)^{{x}} =\mathrm{1} \\…
Question Number 148033 by iloveisrael last updated on 25/Jul/21 Commented by iloveisrael last updated on 25/Jul/21 $$\:\frac{\mathrm{2}^{\mathrm{2020}} +\mathrm{1}}{\mathrm{2}^{\mathrm{2018}} +\mathrm{1}}\:+\:\frac{\mathrm{3}^{\mathrm{2020}} +\mathrm{1}}{\mathrm{3}^{\mathrm{2018}} +\mathrm{1}}\:+\frac{\mathrm{4}^{\mathrm{2020}} +\mathrm{1}}{\mathrm{4}^{\mathrm{2018}} +\mathrm{1}}\:+\frac{\mathrm{5}^{\mathrm{2020}} +\mathrm{1}}{\mathrm{5}^{\mathrm{2018}} +\mathrm{1}}\:+\frac{\mathrm{6}^{\mathrm{2020}}…
Question Number 16944 by Tinkutara last updated on 28/Jun/17 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{digits}\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{number}\:\mathrm{2}^{\mathrm{2005}} \:×\:\mathrm{5}^{\mathrm{2000}} \:\mathrm{when}\:\mathrm{written}\:\mathrm{in} \\ $$$$\mathrm{full}. \\ $$ Commented by RasheedSoomro last updated on 28/Jun/17…
Question Number 148011 by puissant last updated on 25/Jul/21 $$\mathrm{2}^{\left(\frac{\mathrm{x}}{\mathrm{2}}\right)} −\mathrm{1}=\mathrm{3}^{\mathrm{x}} \\ $$ Commented by mr W last updated on 25/Jul/21 $${x}>\mathrm{0}:\: \\ $$$$\mathrm{3}^{{x}} >\left(\sqrt{\mathrm{2}}\right)^{{x}}…
Question Number 147988 by puissant last updated on 24/Jul/21 $$\mathrm{La}\:\mathrm{somme}\:\mathrm{des}\:\mathrm{n}\:\mathrm{premiers}\:\mathrm{termes}\:\mathrm{d}'\mathrm{une} \\ $$$$\mathrm{s}\acute {\mathrm{e}rie}\:\mathrm{est}\:\mathrm{donn}\acute {\mathrm{e}}\:\mathrm{par}\:\mathrm{S}_{\mathrm{n}} =\mathrm{5n}^{\mathrm{2}} +\mathrm{2n}\:\mathrm{le}\: \\ $$$$\mathrm{n}−\mathrm{ieme}\:\mathrm{terme}\:\mathrm{de}\:\mathrm{cette}\:\mathrm{serie}\:\mathrm{est}: \\ $$ Answered by Olaf_Thorendsen last updated…