Question

The value of $S=2^{1/4}\cdot 4^{1/8}\cdot 8^{1/16}\cdot {16}^{1/32}\dots \infty$, is

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

Answer 1

Answer: (A)

$2$

$S=2^{1/4}\cdot 4^{1/8}\cdot 8^{1/16}\cdot {16}^{1/32}\dots \infty$

Let $S=2^y=2^{\frac{1}{4}+\frac{2}{8}+\frac{3}{16}\cdots \infty }$

Now, lets solve the series in the power of $2$.
$y=\frac{1}{4}+\frac{2}{8}+\frac{3}{16}\cdots \infty$ ...(a)

Divide by $2$
$\frac{y}{2}=0+\frac{1}{8}+\frac{2}{16}\cdots \infty$, ...(b)
By subtracting eqn (b) from eqn (a), we get
$\frac{y}{2}=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\cdots \infty$

It is of the form of an infinite G.P., with first term $a=\frac{1}{4}$ and common ratio $r=\frac{1}{2}$.

$\therefore$ The sum of this infinite G.P $=\frac{a}{1-r}$
$\frac{y}{2}=\frac{1}{4}\times \frac{1}{(1-\frac{1}{2})}$

$\frac{y}{2}=\frac{1}{4}\times \frac{1}{\left(\frac{1}{2}\right)}$

$y=\frac{1}{2}\times 2=1$

So, $S=2^1=2$
Hence, option A is the correct answer.