Question

If x = $\sqrt{1+\sqrt{1+\sqrt{1+.....toinfinity}}},$ then x =

• A. $\frac{1+\sqrt{5}}{2}$
• B. $\frac{1-\sqrt{5}}{2}$
• C. $\frac{1\pm \sqrt{5}}{2}$
• D. None of these

Answer 1

The correct option is A $\frac{1+\sqrt{5}}{2}$

x = $\sqrt{1+\sqrt{1+\sqrt{1+.....}}},$ to $\infty$
We have $x=\sqrt{1+x}$
$\Rightarrow x^2=1+x\Rightarrow x^2-x-1=0$
$\Rightarrow x=\frac{1\pm \sqrt{1+4}}{2}=\frac{1\pm \sqrt{5}}{2}$
As $x>0,$ We get x =$\frac{1+\sqrt{5}}{2}$