Q. If $x = \sqrt{1 + \sqrt{1 + \sqrt{1 + \ldots\infty}}},\$then $x$ is equal to
A) $\frac{1 + \sqrt{5}}{2}$
B) $\frac{1 - \sqrt{5}}{2}$
C) $\frac{1 \pm \sqrt{5}}{2}$
D) None of these
Explanation
(a)
We have, $x = \sqrt{1 + \sqrt{1 + \sqrt{1 + \ldots\infty}}}$
$\Rightarrow x = \sqrt{1 + x}$
$\Rightarrow x^{2} = 1 + x\ \Rightarrow x^{2} - x - 1 = 0$
$\Rightarrow x = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2}$
As $x > 0$, we take only $x = \frac{1 + \sqrt{5}}{2}$.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.