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Question Number 119177 by help last updated on 22/Oct/20

Answered by Dwaipayan Shikari last updated on 22/Oct/20

$$\underset{k=0}{\sum ^{\mathrm{\infty }}}{\left(\frac{4}{3}\right)}^{-k}=1+\frac{3}{4}+\frac{3^2}{4^2}+...=\frac{1}{1-\frac{3}{4}}=4$$

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