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Question Number 143596 by ZiYangLee last updated on 16/Jun/21

$$\mathrm{Given}\mathrm{that}\mathrm{the}\mathrm{series}$$$$\frac{x}{3^{}}+\frac{y}{3^2}+\frac{x}{3^3}+\frac{y}{3^4}+\dots +\frac{x}{3^{n-1}}+\frac{y}{3^n}=8,x,y\in \mathbb{R}.$$$$\mathrm{Find}\mathrm{the}\mathrm{value}\mathrm{of}3x+y.$$

Answered by bobhans last updated on 16/Jun/21

$$S_1=x(\frac{1}{3}+\frac{1}{3^3}+\frac{1}{3^5}+...)=x\left(\frac{1/3}{8/9}\right)=\frac{3x}{8}$$$$S_2=y(\frac{1}{3^2}+\frac{1}{3^4}+\frac{1}{3^6}+...)=y\left(\frac{1/9}{8/9}\right)=\frac{y}{8}$$$$\Rightarrow \frac{3x+y}{8}=8\Rightarrow 3x+y=64$$

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