Question

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

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

Answer 1

The correct option is C

$64$

Explanation for correct option

Given series $\left(32\right){\left(32\right)}^{\frac{1}{6}}{\left(32\right)}^{\frac{1}{36}}...\infty$ can also be written as ${\left(32\right)}^{\left[1+\frac{1}{6}+\frac{1}{36}+...\infty \right]}$

Here, $1+\frac{1}{6}+\frac{1}{36}+...\infty$is an infinite G.P. with first term $\mathrm{a}=1$ and common ratio $\mathrm{r}=\frac{1}{6}$

Sum of infinite terms of G.P. $=\frac{\mathrm{a}}{1-\mathrm{r}}$

So, $1+\frac{1}{6}+\frac{1}{36}+...\infty =\frac{1}{1-{\displaystyle \frac{1}{6}}}$

$\Rightarrow$$1+\frac{1}{6}+\frac{1}{36}+...\infty =\frac{1}{{\displaystyle \frac{6-1}{6}}}$

$\Rightarrow$$1+\frac{1}{6}+\frac{1}{36}+...\infty =\frac{6}{{\displaystyle 5}}$

Thus, ${\left(32\right)}^{\left[1+\frac{1}{6}+\frac{1}{36}+...\infty \right]}={32}^{\left[\frac{6}{5}\right]}$

$=\left(2^{5\times \frac{6}{5}}\right)$ $\left[\because {\left({\mathrm{a}}^{\mathrm{m}}\right)}^{\mathrm{n}}=a^{m\times n}\right]$

$={\left(2\right)}^6$

$\Rightarrow$ ${\left(32\right)}^{\left[1+\frac{1}{6}+\frac{1}{36}+...\infty \right]}=64$

Hence, the correct option is option(C) i.e. $64$