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Question Number 165210 by mkam last updated on 27/Jan/22

$$\underset{\mathit{n}=4}{\sum ^{\mathrm{\infty }}}2\times {\left(\frac{3}{4}\right)}^{\mathit{n}+3}$$

Commented by cortano1 last updated on 27/Jan/22

$$2\times [{\left(\frac{3}{4}\right)}^7+{\left(\frac{3}{4}\right)}^8+...]=2\times \frac{{\left(\frac{3}{4}\right)}^7}{\left(\frac{1}{4}\right)}$$$$=8\times {\left(\frac{3}{4}\right)}^7$$

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