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