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Question-188926




Question Number 188926 by Math453266 last updated on 09/Mar/23
Answered by qaz last updated on 10/Mar/23
x_n =1+C_n ^3 +C_n ^6 +...  y_n =C_n ^1 +C_n ^4 +C_n ^7 +...  z_n =C_n ^2 +C_n ^5 +C_n ^8 +...  1+ω+ω^2 =0  ⇒ { ((x_n +y_n +z_n =2^n )),((x_n +ωy_n +ω^2 z_n =(1+ω)^n =(−ω^2 )^n =cos ((nπ)/3)+isin ((nπ)/3))),((x_n +ω^2 y_n +ωz_n =(1+ω^2 )^n =(−ω)^n =cos ((nπ)/4)−isin ((nπ)/3))) :}  ⇒(a),(b),(c)
$${x}_{{n}} =\mathrm{1}+{C}_{{n}} ^{\mathrm{3}} +{C}_{{n}} ^{\mathrm{6}} +… \\ $$$${y}_{{n}} ={C}_{{n}} ^{\mathrm{1}} +{C}_{{n}} ^{\mathrm{4}} +{C}_{{n}} ^{\mathrm{7}} +… \\ $$$${z}_{{n}} ={C}_{{n}} ^{\mathrm{2}} +{C}_{{n}} ^{\mathrm{5}} +{C}_{{n}} ^{\mathrm{8}} +… \\ $$$$\mathrm{1}+\omega+\omega^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\begin{cases}{{x}_{{n}} +{y}_{{n}} +{z}_{{n}} =\mathrm{2}^{{n}} }\\{{x}_{{n}} +\omega{y}_{{n}} +\omega^{\mathrm{2}} {z}_{{n}} =\left(\mathrm{1}+\omega\right)^{{n}} =\left(−\omega^{\mathrm{2}} \right)^{{n}} =\mathrm{cos}\:\frac{{n}\pi}{\mathrm{3}}+{i}\mathrm{sin}\:\frac{{n}\pi}{\mathrm{3}}}\\{{x}_{{n}} +\omega^{\mathrm{2}} {y}_{{n}} +\omega{z}_{{n}} =\left(\mathrm{1}+\omega^{\mathrm{2}} \right)^{{n}} =\left(−\omega\right)^{{n}} =\mathrm{cos}\:\frac{{n}\pi}{\mathrm{4}}−{i}\mathrm{sin}\:\frac{{n}\pi}{\mathrm{3}}}\end{cases} \\ $$$$\Rightarrow\left({a}\right),\left({b}\right),\left({c}\right) \\ $$

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