Question

$2^{\frac{1}{4}}\times 4^{\frac{1}{8}}\times 8^{\frac{1}{16}}...$ to $\infty$ is equal to

• A. $1$
• B. $2$
• C. $\frac{3}{2}$
• D. None of these

Answer 1

The correct option is B $2$

Let $A=2^{\frac{1}{4}}.4^{\frac{1}{8}}.8^{\frac{1}{16}}...=2^{\frac{1}{2^2}}.2^{\frac{2}{2^3}}.2^{\frac{3}{2^4}}...=2^{\frac{1}{2^2}+\frac{2}{2^3}+\frac{3}{2^4}...}$

Now let $S=\frac{1}{2^2}+\frac{2}{2^3}+\frac{3}{2^4}+...$ ...(1)

$\Rightarrow \frac{1}{2}S=\frac{1}{2^3}+\frac{2}{24}+\frac{3}{25}+...$ ...(2)

Now $\left(1\right)-\left(2\right)$, we get
$\Rightarrow \frac{1}{2}S=\frac{1}{2^2}+\frac{2}{2^3}+\frac{3}{2^4}+...\Rightarrow \frac{1}{2}S=\frac{\frac{1}{2^2}}{1-\frac{1}{2}}$

$\Rightarrow \frac{1}{2}S=\frac{\frac{1}{4}}{\frac{1}{2}}=\frac{1}{2}\Rightarrow S=1\therefore A=2^1=2$