Question

The product of infinite terms in $x^{\frac{1}{2}}.x^{\frac{1}{4}}.x^{\frac{1}{8}}...........\infty$ is

• A. 0
• B. infinity
• C. 1
• D. x

Answer 1

The correct option is D

x

$x^{\frac{1}{2}}.x^{\frac{1}{4}}.x^{\frac{1}{8}}...........\infty$
$=x^{(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+.....+\infty )}$
$=x^{(\frac{1}{2}+{\left(\frac{1}{2}\right)}^2+{\left(\frac{1}{2}\right)}^3+.....+\infty )}$
Consider
$\frac{1}{2}+{\left(\frac{1}{2}\right)}^2+{\left(\frac{1}{2}\right)}^3+.....+\infty S_{\infty }=\frac{a}{1-r}=\frac{\frac{1}{2}}{1-\frac{1}{2}}=1\Rightarrow x^{(\frac{1}{2}+{\left(\frac{1}{2}\right)}^2+.........+\infty )}=x^1=x$