Problem - 3861
Show that $$x+n=\sqrt{n^2 + x\sqrt{n^2+(x+n)\sqrt{n^2+(x+2n)\sqrt{\cdots}}}}$$
This equation is named after the Indian mathematician Srinivasa Ramanujan. It holds because
\begin{align}
& x + n \\
=\quad& \sqrt{(n+x)^2}\\
=\quad& \sqrt{n^2 + x (x + n + n)}\\
=\quad& \sqrt{n^2 + x\sqrt{(n+(x+n))^2}}\\
=\quad& \sqrt{n^2 + x\sqrt{n^2 + (x+n)(x+n + 2n)}}\\
=\quad& \sqrt{n^2 + x\sqrt{n^2 + (x+n)\sqrt{(n+(x+2n))^2}}}\\
=\quad& \cdots
\end{align}