A Radical Reciprocation

$\begin{aligned} S & = \frac 1{\sqrt 1+\sqrt 2} + \frac 1{\sqrt 2+\sqrt 3} + \frac 1{\sqrt 3+\sqrt 4} + ... + \frac 1{\sqrt n+\sqrt {n+1}} \\ & = \small \frac {-\sqrt 1+\sqrt 2}{(\sqrt 1+\sqrt 2)(-\sqrt 1+\sqrt 2)} + \frac {-\sqrt 2+\sqrt 3}{(\sqrt 2+\sqrt 3)(-\sqrt 2+\sqrt 3)} + \frac {-\sqrt 3+\sqrt 4}{(\sqrt 3+\sqrt 4)(-\sqrt 3+\sqrt 4)} + ... +\frac {-\sqrt n+\sqrt {n+1}}{(\sqrt n+\sqrt {n+1})(-\sqrt n+\sqrt {n+1})} \\ & = \frac {-\sqrt 1+\sqrt 2}1 + \frac {-\sqrt 2+\sqrt 3}1 + \frac {-\sqrt 3+\sqrt 4}1 + ... +\frac {-\sqrt n+\sqrt {n+1}}1 \\ & = -\sqrt 1+\sqrt {n+1} \\ \implies S & = \sqrt {n+1} - 1 \end{aligned}$

So $S$ is rational when $\sqrt {n+1}$ is a positive integer or $n+1$ , a perfect square $\ge 4$ . Therefore, the sum of values of required $n$ is as follows:

$\begin{aligned} S_n & = \sum_{k=2}^{10} (k^2-1) \\ & = \frac {10(11)(21)}6 - 1 - 9 \\ & = \boxed{375} \end{aligned}$

Note that $\frac{1}{\sqrt{n}+\sqrt{n+1}} = \frac{\sqrt{n+1}-\sqrt{n}}{(\sqrt{n}+\sqrt{n+1})(\sqrt{n+1}-\sqrt{n})}$ $= \sqrt{n+1} - \sqrt{n}$

So the series is equivalent to $\sqrt{n+1} -1$ and is rational iff $n+1=x^{2}$ for $x^{2}: 1 \leq x^{2} \leq 100$

The solutions are then $[ 3,8,15,24,35,48,63,80,99]$ which gives a sum of $375$