On $3+\sqrt{11+\sqrt{11+\sqrt{11+\sqrt{11+\dots}}}}=\phi^4$ and friends Let $\phi$ be the golden ratio. We know it has a beautiful infinite nested radical,
$$\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\dots}}}}=\phi$$
However, it is also the case that,
$$3+\sqrt{11+\sqrt{11+\sqrt{11+\sqrt{11+\dots}}}}=\phi^4$$
$$5+\sqrt{31+\sqrt{31+\sqrt{31+\sqrt{31+\dots}}}}=\phi^5$$
$$\tfrac{17}2+\sqrt{\tfrac{319}4+\sqrt{\tfrac{319}4+\sqrt{\tfrac{319}4+\sqrt{\tfrac{319}4+\dots}}}}=\phi^6$$

Q: How do we show that, in general
$$a_n+\sqrt{b_n+\sqrt{b_n+\sqrt{b_n+\sqrt{b_n+\dots}}}}=\phi^n$$
where,
$$a_n = \frac{L_n-1}2,\quad b_n = \frac{5F_n^2-1}4$$
with Lucas numbers $L_n$ and Fibonacci numbers $F_n$?

If we plug in the given formulas we get the famous formula. We note:
$$\sqrt{b_n+x}=x \Rightarrow x=\frac{1+\sqrt{1+4b_n}}{2}=\frac{1+F_n\sqrt{5}}{2};$$
Hence:
$$\frac{L_n-1}{2}+\frac{1+F_n\sqrt{5}}{2}=\frac{L_n+F_n\sqrt{5}}{2}=\phi^n.$$
This is a famous formula that relates the two sequences. See: https://en.m.wikipedia.org/wiki/Lucas_number
