3
$\begingroup$

$\frac{\sqrt{31+\sqrt{31+\sqrt{31+ \cdots}}}}{\sqrt{1+\sqrt{1+ \sqrt{1+ \cdots}}}}=a-\sqrt b$ where $a,b$ are natural numbers. Find the value of $a+b$.

I am not able to proceed with solving this question as I have no idea as to how I can calculate $\frac{\sqrt{31+\sqrt{31+\sqrt{31+ \cdots}}}}{\sqrt{1+\sqrt{1+ \sqrt{1+ \cdots}}}}$. A small hint would do.

$\endgroup$
  • 2
    $\begingroup$ If $x=\sqrt{1+\cdots}$ then (you'll have to prove that) $x=\sqrt{1+x}$, and similarly for the other continued radical. $\endgroup$ – J. M. isn't a mathematician Dec 15 '16 at 11:21
  • 2
    $\begingroup$ Evaluate $y$ for $y=\sqrt{x+y}$ and $x\in\{31;1\}$ . The result is $(a;b)=(6;5)$ . $\endgroup$ – user90369 Dec 15 '16 at 11:29
  • $\begingroup$ Let me guess: from Brilliant.org? $\endgroup$ – Bart Michels Dec 17 '16 at 0:30
6
$\begingroup$

To find $\sqrt{a+\sqrt{a+\cdots}} $, solve the equation

$x = \sqrt{a+x}$

The solution of $\sqrt{31+\sqrt{31+\cdots}}$ gives us $x = \frac{1\pm 5\sqrt{5}}{2}$.

The solution of $\sqrt{1+\sqrt{1+\cdots}}$ gives us $y = \frac{1\pm \sqrt{5}}{2}$.

Thus, we have $$\frac{x}{y} = 6-\sqrt{5}$$ Hence, $a=6, b=5 \Rightarrow a+b = 11$. Hope it helps.

| cite | improve this answer | |
$\endgroup$
4
$\begingroup$

Hint: To find

$$\sqrt{a+\sqrt{a+\cdots}} $$

solve the equation

$$x = \sqrt{a+x}$$

As for the answer, I get $11$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.