Nested again

In problems like these, $f(x)=x+n+a$ .

Also, $f(x)=\sqrt{ax+(n+a)^2+x\sqrt{ax+(n+a)^2+(x+n)\sqrt{ax+...}}}$ , and it is fairly obvious that a=0, n=1, and x=3.

So just add those to get 4, which is the answer.

Thanks Ramanujan!

x²=1+x²-1.......(i)

x²=1+(x+1)(x-1)

Now Further,

x+1=√(x+1)²

x+1=√1+(x+2)(x) (using (i) )

Again

x+2=√1+(x+3)(x+1)

Finally

x²=1+(x-1)√1+(x) √1+(x+3)(x+1)... and so on

x=√1+(x-1)√1+(x) √1+(x+1)(x+3)...

Now according to the given equation put x=4 :

4=√1+(4-1)√1+(4) √1+(4+1)...

4=√1+3√1+4 √1+5...

Solution edited to avoid inappropriate comments on a public forum

[x+n]^2 = n^2 + 2nx + x^2 = n^2 + x [(x+n) + n] [x+n] = sqrt(n^2 + x [(x+n)+n]) replace x+n with its definition. x+n = sqrt(n^2 + x sqrt(n^2 + (x+n)[(x+2n)+n])) x+n = sqrt(n^2 + x sqrt(n^2 + (x+n) sqrt(n^2 + (x+2n) sqrt(...)...) set n = 1, x = 3 to get 3+1 = 4 = sqrt(1+3sqrt(1+(4sqrt(1+(5sqrt())))

so the answer is 4

$Ramanujan\ identity\ \ \ \large\ 3= \sqrt{1 + 2\sqrt{1 + 3\sqrt{1 + 4\sqrt{1 + 5\sqrt{1 + 6\sqrt{1 + \ldots}}}}}}\\ Squaring\ both\ sides \large\ \ \ 9=1+2\sqrt{1 + 3\sqrt{1 + 4\sqrt{1 + 5\sqrt{1 + 6\sqrt{1 + \ldots}}}}}\\ \therefore\ {\Huge \color{#D61F06}{4}}\ \ =\large \sqrt{1 + 3\sqrt{1 + 4\sqrt{1 + 5\sqrt{1 + 6\sqrt{1 + \ldots}}}}}\\$

One key to answer this question is to know the "great" Ramanujan identity. Beautiful identity indeed.

The crux move to this problem is that, n^2=1+(n-1)(n+1) So, n=sqrt(1+(n-1)(n+1)) Then you can do the same thing with (n+1)

function rr(n){

var last=n Math.sqrt(2+n); while (n>3){ last=(n-1) Math.sqrt(1+last); n--;

} return Math.sqrt(1+last); }