# Nested Square Roots $5^0+\sqrt{5^1+\sqrt{5^2+\sqrt{5^4+\sqrt\dots}}}$

How would one go about computing the value of $X$, where

$X=5^0+ \sqrt{5^1+\sqrt{5^2+\sqrt{5^4+\sqrt{5^8+\sqrt{5^{16}+\sqrt{5^{32}+\dots}}}}}}$

I have tried the standard way of squaring then trying some trick, but nothing is working. I have also looked at some nested radical previous results, but none seem to be of the variety of this problem. Can anyone come up with the answer? Thank

The trick is to pull out a $\sqrt{5}$ factor from the second term:
$$\frac{\sqrt{5^1+ \sqrt{5^2 + \sqrt{5^4 + \sqrt{5^8 + \cdots}}}}}{\sqrt{5}} = \sqrt{1 + \sqrt{1 + \sqrt{1 + \sqrt{1 + \cdots}}}},$$ which I call $Y$ for convenience. To see why this is true, observe that \begin{align*} \frac{\sqrt{5^1+x}}{\sqrt{5}} = \sqrt{1+\frac{x}{5^1}} \\ \frac{\sqrt{5^2+x}}{5^1} = \sqrt{1+\frac{x}{5^2}} \\ \frac{\sqrt{5^4+x}}{5^2} = \sqrt{1+\frac{x}{5^4}} \end{align*} and so on. Applying these repeatedly inside the nested radicals gives the claim. (The wording of this explanation has been inspired mostly by a comment of @J.M. below.)
Now, it only remains to compute $Y$ and $X$. By squaring, we get $$Y^2 = 1 + Y \ \ \ \ \implies Y = \frac{\sqrt{5}+1}{2},$$ discarding the negative root. Plugging this value in the definition of $X$, we get: $$X = 1 + \sqrt{5} Y = 1 + \frac{5+\sqrt{5}}{2} = \frac{7+\sqrt{5}}{2} .$$
• @Fool: you mean the nested radical? Note that $\sqrt{5+x}/\sqrt 5=\sqrt{1+x/5}$; just keep applying that as you go deeper. – J. M. is a poor mathematician Aug 31 '11 at 20:12
• $Y^2=1+Y$ has $Y=\infty$ as solution, as well as the one you found... – GEdgar Aug 31 '11 at 21:09