Mathematical Enigma - XIX

From ramanujan Nested Radicals Formula we get :

A simple successive elevation to square and subtraction/division leads to

For a demonstration of the Nested Radicals Formula, consider this :

Let $f(x) = x+1$ . Then

$\begin{aligned} f(x) & = x+1 = \sqrt{(x+1)^2} = \sqrt{x^2+2x+1} = \sqrt{1+x(x+2)} \\ & = \sqrt{1+xf(x+1)} \\ & = \sqrt{1+x\sqrt{1+(x+1)f(x+2)}} \\ & = \sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)f(x+3)}}} \\ & = \sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+(x+3)\sqrt{1+\cdots}}}}} \\ \implies f(5) & = \sqrt{1+5\sqrt{1+6\sqrt{1+7\sqrt{1+8\sqrt {1+ \cdots}}}}} \\ & = 5 + 1 = \boxed 6 \end{aligned}$

Substitution from ramanujan series and i got the answer is 6