Question

The value of $4^{\frac{1}{3}}.4^{\frac{1}{9}}.4^{\frac{1}{27}}..........\infty$ is

• A. $2$
• B. $3$
• C. $4$
• D. $9$

Answer 1

The correct option is A

$2$

Explanation for the correct answer:

By using the properties of indices we can write

$4^{\frac{1}{3}}.4^{\frac{1}{9}}.4^{\frac{1}{27}}..........\infty =4^{\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+......\infty \right)}$

The exponent is the sum of an infinite geometric progression with the first term as $\frac{1}{3}$ and common ratio as $\frac{1}{3}$

The sum of an infinite geometric progression is given as

$S_{\infty }=\frac{a}{1-r}$

We have , $a=r=\frac{1}{3}$

$\Rightarrow$ $S_{\infty }=\frac{{\displaystyle \frac{1}{3}}}{1-{\displaystyle \frac{1}{3}}}$

$=\frac{{\displaystyle \frac{1}{3}}}{{\displaystyle \frac{2}{3}}}$

$\Rightarrow$ $S_{\infty }=\frac{1}{2}$

$\Rightarrow$ $4^{\frac{1}{3}}.4^{\frac{1}{9}}.4^{\frac{1}{27}}..........\infty =4^{\frac{1}{2}}$

$\Rightarrow$$4^{\frac{1}{3}}.4^{\frac{1}{9}}.4^{\frac{1}{27}}..........\infty =2$

Hence, the value of $4^{\frac{1}{3}}.4^{\frac{1}{9}}.4^{\frac{1}{27}}..........\infty$ is $2$.

Hence, option $\left(A\right)$ is the correct answer.