Question

The value of $2^{\frac{1}{4}}.4^{\frac{1}{8}}.8^{\frac{1}{16}}........\infty$ is

• A. $1$
• B. $2$
• C. $\frac{3}{2}$
• D. $4$

Answer 1

The correct option is B

$2$

Explanation for the correct answer:

Let $S=2^{\frac{1}{4}}.4^{\frac{1}{8}}.8^{\frac{1}{16}}........\infty$

Rewrite all terms as powers of $2$ we get,

$S=2^{\frac{1}{4}}.2^{\frac{2}{8}}.2^{\frac{3}{16}}.........\infty$

$\Rightarrow$ $S=2^{\left(\frac{1}{4}+\frac{2}{8}+\frac{3}{16}+....\infty \right)}=2^k$ (say)

$\Rightarrow$ $k=\left(\frac{1}{4}+\frac{2}{8}+\frac{3}{16}+\frac{4}{32}....\infty \right)$ $...\left(i\right)$

Dividing by $2$ we get

$\Rightarrow$ $\frac{k}{2}=\left(0+\frac{1}{8}+\frac{2}{16}+\frac{3}{132}+....\infty \right)$ $...\left(ii\right)$

Subtracting $\left(ii\right)$ from $\left(i\right)$ we get

$\Rightarrow$ $\frac{k}{2}=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+.....\infty$

This sequence is an infinite geometric progression with $a=\frac{1}{4},r=\frac{1}{2}$

The sum of this infinite geometric progression is given as

$\Rightarrow$ $\frac{k}{2}=\frac{a}{1-r}$

$\Rightarrow$ $\frac{k}{2}=\frac{{\displaystyle \frac{1}{4}}}{1-{\displaystyle \frac{1}{2}}}$

$\Rightarrow$ $\frac{k}{2}=\frac{{\displaystyle \frac{1}{4}}}{{\displaystyle \frac{1}{2}}}$

$\Rightarrow$ $\frac{k}{2}=\frac{1}{2}$

$\Rightarrow$ $k=1$

$\Rightarrow$ $S=2^k=2^1=2$

Hence, the value of $2^{\frac{1}{4}}.4^{\frac{1}{8}}.8^{\frac{1}{16}}........\infty$ is $2$.

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