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

\underset{n = 4}{\sum^{\infty}} 2 \times {(\frac{3}{4})}^{n + 3}

Commented by cortano1 last updated on 27/Jan/22

2 \times [{(\frac{3}{4})}^{7} + {(\frac{3}{4})}^{8} + \dots] = 2 \times \frac{{(\frac{3}{4})}^{7}}{(\frac{1}{4})}

= 8 \times {(\frac{3}{4})}^{7}