Question Number 58409 by Tawa1 last updated on 22/Apr/19 $$\mathrm{Prove}\:\mathrm{without}\:\mathrm{mathematical}\:\mathrm{induction}\:\mathrm{that}\:\mathrm{the}\: \\ $$$$\mathrm{expression}\:\:\:\left(\mathrm{1}\:+\:\sqrt{\mathrm{2}}\right)^{\mathrm{2n}} \:+\:\left(\mathrm{1}\:−\:\sqrt{\mathrm{2}}\right)^{\mathrm{2n}} \:\:\mathrm{is}\:\mathrm{even}\:\mathrm{for}\:\mathrm{every} \\ $$$$\mathrm{natural}\:\mathrm{number}\:\:\mathrm{n}. \\ $$ Commented by maxmathsup by imad last updated…
Question Number 58410 by Tawa1 last updated on 22/Apr/19 $$\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{cube}\:\mathrm{of}\:\mathrm{three}\:\mathrm{consecutive} \\ $$$$\mathrm{number}\:\mathrm{gives}\:\mathrm{a}\:\mathrm{multiple}\:\mathrm{of}\:\:\mathrm{9}. \\ $$ Answered by mr W last updated on 22/Apr/19 $${S}=\left({n}−\mathrm{1}\right)^{\mathrm{3}} +{n}^{\mathrm{3}} +\left({n}+\mathrm{1}\right)^{\mathrm{3}}…
Question Number 189473 by Spillover last updated on 17/Mar/23 Answered by a.lgnaoui last updated on 17/Mar/23 $${x}=\mathrm{1}\:\:\:{f}\left(\mathrm{1}\right)=\left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{3}} =\frac{\mathrm{27}}{\mathrm{64}} \\ $$$${x}=\mathrm{2}\:\:{f}\left(\mathrm{2}\right)=\left(\frac{\mathrm{3}}{\mathrm{8}}\right)^{\mathrm{2}} =\frac{\mathrm{9}}{\mathrm{64}} \\ $$$${x}=\mathrm{3}\:\:{f}\left(\mathrm{3}\right)=\left(\frac{\mathrm{3}}{\mathrm{12}}\right)^{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{4}}=\frac{\mathrm{16}}{\mathrm{64}} \\…
Question Number 58402 by rahul 19 last updated on 22/Apr/19 $${The}\:{imaginary}\:{part}\:{of}\:\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}{i}\right)^{\mathrm{10}} {is}\:? \\ $$ Commented by maxmathsup by imad last updated on 23/Apr/19 $${let}\:{Z}\:=\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}{i}\right)^{\mathrm{10}} \:\Rightarrow{Z}\:=\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{10}}…
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Question Number 58390 by MJS last updated on 22/Apr/19 $$\mathrm{write}\:\mathrm{without}\:\mathrm{roots}\:\mathrm{in}\:\mathrm{denominator}\:\mathrm{if}\:\mathrm{possible} \\ $$$$\left(\mathrm{1}\right)\:\frac{\mathrm{1}}{\:\sqrt{{a}}} \\ $$$$\left(\mathrm{2}\right)\:\frac{\mathrm{1}}{\:\sqrt{{a}}+\sqrt{{b}}} \\ $$$$\left(\mathrm{3}\right)\:\frac{\mathrm{1}}{\:\sqrt{{a}}+\sqrt{{b}}+\sqrt{{c}}} \\ $$$$\left(\mathrm{4}\right)\:\frac{\mathrm{1}}{\:\sqrt{{a}}+\sqrt{{b}}+\sqrt{{c}}+\sqrt{{d}}} \\ $$$$\left(\mathrm{5}\right)\:\frac{\mathrm{1}}{\:\sqrt{{a}}+\sqrt{{b}}+\sqrt{{c}}+\sqrt{{d}}+\sqrt{{e}}} \\ $$ Commented by tanmay…
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Question Number 189446 by mathlove last updated on 16/Mar/23 $${x}^{\mathrm{2}} =\sqrt[{\mathrm{5}}]{\mathrm{2}}+{y} \\ $$$${y}^{\mathrm{2}} =\sqrt[{\mathrm{5}}]{\mathrm{2}}+{x} \\ $$$${x}\centerdot{y}=?\:\:\:\:\:\:\:\:\:\:\:{x}\neq{y} \\ $$ Answered by cortano12 last updated on 16/Mar/23…
Question Number 123894 by john_santu last updated on 29/Nov/20 Answered by mr W last updated on 29/Nov/20 $${nn}!=\left({n}+\mathrm{1}\right){n}!−{n}!=\left({n}+\mathrm{1}\right)!−{n}! \\ $$$$\Rightarrow{a}_{{n}} =\left({n}+\mathrm{1}\right)!−\mathrm{1} \\ $$$$\frac{{n}}{\left({n}+\mathrm{1}\right)!}=\frac{{n}+\mathrm{1}}{\left({n}+\mathrm{1}\right)!}−\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)!}=\frac{\mathrm{1}}{{n}!}−\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)!} \\ $$$$\Rightarrow{b}_{{n}}…
Question Number 123886 by Ar Brandon last updated on 29/Nov/20 $$\mathrm{Let}\:\overset{−} {{l}z}+{l}\overset{−} {{z}}+{m}=\mathrm{0}\:\mathrm{be}\:\mathrm{a}\:\mathrm{straight}\:\mathrm{line}\:\mathrm{in}\:\mathrm{the}\:\mathrm{complex}\:\mathrm{plane} \\ $$$$\mathrm{and}\:{P}\left({z}_{\mathrm{0}} \right)\:\mathrm{be}\:\mathrm{a}\:\mathrm{point}\:\mathrm{in}\:\mathrm{the}\:\mathrm{plane}.\:\mathrm{Then}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{line}\:\mathrm{passing}\:\mathrm{through}\:{P}\left({z}_{\mathrm{0}} \right)\:\mathrm{and}\:\mathrm{perpendicular} \\ $$$$\mathrm{to}\:\mathrm{the}\:\mathrm{given}\:\mathrm{line}\:\mathrm{is}\:\_\_\_ \\ $$ Answered by…