Question Number 143596 by ZiYangLee last updated on 16/Jun/21
$$\mathrm{Given}\:\mathrm{that}\:\mathrm{the}\:\mathrm{series} \\ $$$$\frac{{x}}{\mathrm{3}^{} }+\frac{{y}}{\mathrm{3}^{\mathrm{2}} }+\frac{{x}}{\mathrm{3}^{\mathrm{3}} }+\frac{{y}}{\mathrm{3}^{\mathrm{4}} }+\ldots+\frac{{x}}{\mathrm{3}^{{n}−\mathrm{1}} }+\frac{{y}}{\mathrm{3}^{{n}} }=\mathrm{8},\:{x},{y}\in\mathbb{R}. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{3}{x}+{y}. \\ $$
Answered by bobhans last updated on 16/Jun/21
$${S}_{\mathrm{1}} ={x}\left(\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{5}} }+...\right)={x}\left(\frac{\mathrm{1}/\mathrm{3}}{\mathrm{8}/\mathrm{9}}\right)=\frac{\mathrm{3}{x}}{\mathrm{8}} \\ $$$${S}_{\mathrm{2}} ={y}\left(\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{4}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{6}} }+...\right)={y}\left(\frac{\mathrm{1}/\mathrm{9}}{\mathrm{8}/\mathrm{9}}\right)=\frac{{y}}{\mathrm{8}} \\ $$$$\Rightarrow\:\frac{\mathrm{3}{x}+{y}}{\mathrm{8}}=\mathrm{8}\Rightarrow\mathrm{3}{x}+{y}=\mathrm{64} \\ $$