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

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(A) Let the given expression be $S = {4^{1/3}} \cdot {4^{1/9}} \cdot {4^{1/27}} \cdots \infty$.Using the law of exponents ${a^m} \cdot {a^n} = {a^{m+n

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$2$

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(A) Let the given expression be $S = {4^{1/3}} \cdot {4^{1/9}} \cdot {4^{1/27}} \cdots \infty$.Using the law of exponents ${a^m} \cdot {a^n} = {a^{m+n}}$,we get:$S = {4^{(1/3 + 1/9 + 1/27 + \cdots \infty)}}$.The exponent is an infinite geometric series with the first term $a = 1/3$ and common ratio $r = 1/3$.The sum of an infinite geometric series is given by $S_{\infty} = \frac{a}{1-r}$.Substituting the values,the sum of the exponent is $\frac{1/3}{1 - 1/3} = \frac{1/3}{2/3} = 1/2$.Therefore,$S = {4^{1/2}} = \sqrt{4} = 2$.

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

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