NESTED!!!

$\color{#3D99F6}{x+n+a=\sqrt{ax+(n+a)^2+x\sqrt{a(x+n)+(n+a)^2+(x+n)\sqrt{a(x+2n)+(n+a)^2+(x+2n)\sqrt{\dotsm}}}}}$ The above formula was discovered by Ramanujan.If we let $\color{#BA33D6}{a=0,n=1}$ ,after simplifying we get: $\color{#3D99F6}{x+1=\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{\dotsm}}}}}$ Our question fits this form,so letting $\color{#20A900}{x=2}$ ,we get: $\color{#3D99F6}{2+1=\boxed{3}=\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{\dotsm}}}}}$

The basic idea is to take advantage of the identity:

(x + k a)² = (a² + (x + (k -1) a) (x + k a + a))

So, starting with k = 1, we have:

(x + a) = √(a² + x(x + a + a)))

which leads to:

(x + a) = √(a² + x√(a² + (x + a)(x + 2a + a))

and so on, so that we have:

(x + a) = √(a² + x√(a² + (x + a)√(a² + (x + 2a)√(a² + (x + 3a)√(a² + ...

Letting x = 2 and a = 1, we have:

2 + 1 = 3 = √( 1 + 2√(1 + 3√(1 + 4√(1 + ...

so there