Question

The infinite product $x^{\frac{1}{2}}.x^{\frac{1}{4}}.x^{\frac{1}{8}}...........$ equals ___.

• A. 0
• B. infinity
• C. 1
• D. x

Answer 1

The correct option is D

x

$x^{\frac{1}{2}}.x^{\frac{1}{4}}.x^{\frac{1}{8}}...=x^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...}$
$=x^{\frac{1}{2}+{\left(\frac{1}{2}\right)}^2+{\left(\frac{1}{2}\right)}^3+...}$

$\text{Let}S=\frac{1}{2}+{\left(\frac{1}{2}\right)}^2+{\left(\frac{1}{2}\right)}^3+...$

Since the sum of infinite terms of a GP of first term $a$ and common ratio $r$ is given by $\frac{a}{1-r},$ we have
$S=\frac{\frac{1}{2}}{1-\frac{1}{2}}=1.\Rightarrow x^{\frac{1}{2}+{\left(\frac{1}{2}\right)}^2+{\left(\frac{1}{2}\right)}^3...}=x^1\Rightarrow x^{\frac{1}{2}}.x^{\frac{1}{4}}.x^{\frac{1}{8}}...=x$