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Question Number 143708 by ZiYangLee last updated on 17/Jun/21

Prove that

\frac{3}{4^{}} + \frac{3^{2}}{4^{2}} + \frac{3^{3}}{4^{3}} + \frac{3^{4}}{4^{4}} + \frac{3^{5}}{4^{5}} + \dots = 3

Answered by som(math1967) last updated on 17/Jun/21

3 {\frac{1}{4} + (\frac{1}{4}) \times (\frac{3}{4}) + \frac{1}{4} \times {(\frac{3}{4})}^{2} + \dots}

= 3 \times \frac{\frac{1}{4}}{1 - \frac{3}{4}} = 3 \times 1 = 3

∵ a + {a r}^{2} + {a r}^{3} + \dots . = \frac{a}{1 - r}

Answered by Dwaipayan Shikari last updated on 17/Jun/21

a + a^{2} + a^{3} + \dots = \frac{a}{1 - a}

\frac{3}{4} + \frac{3^{2}}{4^{2}} + . . = \frac{\frac{3}{4}}{1 - \frac{3}{4}} = 3

Answered by Willson last updated on 17/Jun/21

\sum_{n ⩾ 1} {(\frac{3}{4})}^{n} = \frac{3}{4} \sum_{n ⩾ 0} {(\frac{3}{4})}^{n} = \frac{3}{4} (\frac{1}{1 - \frac{3}{4}}) = 3

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