Question

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

• A. $1$
• B. $2$
• C. $4$
• D. None of these

Answer 1

Answer: (B)

$2$

We have to find the value of $2^{\frac{1}{4}}{.4}^{\frac{1}{8}}{.8}^{\frac{1}{16}}........$

$\Rightarrow 2^{(\frac{1}{2^2}+\frac{2}{2^3}+\frac{3}{2^4}+......)}=2^{\frac{1}{4}(1+\frac{2}{2^1}+\frac{3}{2^2}+......)}$

Observe that it is in the form of arithmetic geometric progression

In arithmetic progression first term is $a=1$ and difference is $d=1$

In geometric progression , the common term is $r=\frac{1}{2}$

Therefore the sum of infinite AGP series $(1+\frac{2}{2^1}+\frac{3}{2^2}+......)$ is $\frac{a}{1-r}+\frac{dr}{(1-r{)}^2}=2+2=4$

So, the value of $2^{(\frac{1}{2^2}+\frac{2}{2^3}+\frac{3}{2^4}+......)}$ is $2^{(\frac{1}{4}\times 4)}=2$

Therefore the correct option is $B$