The value of {4^{1/3}} cdot {4^{1/9}} cdot {4 | vedclass

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(A) Let $S = {4^{1/3}} \cdot {4^{1/9}} \cdot {4^{1/27}} \cdots \infty$. Since the bases are the same,we add the exponents: $S = {4^{(1/3 + 1/9 + 1/27

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$2$

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(A) Let $S = {4^{1/3}} \cdot {4^{1/9}} \cdot {4^{1/27}} \cdots \infty$. Since the bases are the same,we add the exponents: $S = {4^{(1/3 + 1/9 + 1/27 + \cdots \infty)}}$. The exponent is an infinite geometric series with first term $a = 1/3$ and common ratio $r = 1/3$. The sum of an infinite geometric series is given by $S_{\infty} = \frac{a}{1 - r}$. $S_{\infty} = \frac{1/3}{1 - 1/3} = \frac{1/3}{2/3} = \frac{1}{2}$. Therefore,$S = {4^{1/2}} = 2$.

The value of ${4^{1/3}} \cdot {4^{1/9}} \cdot {4^{1/27}} \cdots \infty$ is

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