Question

The product $32{\left(32\right)}^{\frac{1}{6}}{\left(32\right)}^{\frac{1}{36}}...........\infty$ is

• A. $16$
• B. $32$
• C. $64$
• D. $0$

Answer 1

The correct option is C

$64$

Explanation for the correct option:

$$\begin{gathered} \Rightarrow 32{\left(32\right)}^{\frac{1}{6}}{\left(32\right)}^{\frac{1}{36}}...........\infty \\ \Rightarrow {32}^{1+\frac{1}{6}+\frac{1}{36}+.......\infty } \end{gathered}$$

Now, $1,\frac{1}{6},\frac{1}{36},.........\infty$ is in geometric progression with the first term $a=1$ and common ratio $r=\frac{1}{6}$

So, the sum of infinite geometric progression $\mathbf{=}\frac{\mathbf{a}}{\mathbf{1}\mathbf{-}\mathbf{r}}$

sum of infinite geometric progression $1,\frac{1}{6},\frac{1}{36},.........\infty$$=\frac{1}{1-{\displaystyle \frac{1}{6}}}$

$$\begin{gathered} =\frac{1}{{\displaystyle \frac{5}{6}}} \\ =\frac{6}{5} \end{gathered}$$

$$\begin{gathered} \Rightarrow 32{\left(32\right)}^{\frac{1}{6}}{\left(32\right)}^{\frac{1}{36}}...........\infty ={\left(32\right)}^{\frac{6}{5}} \\ \Rightarrow ={\left(2^5\right)}^{\frac{6}{5}} \\ \Rightarrow =2^6 \\ \Rightarrow =64 \end{gathered}$$

Hence, option C is correct.