<s>[INST] Series of Nested Radicals 
I can't seem to find a way, squaring the expression would make more terms and would make it harder, I guess there must be something to do with the first and last terms as they sum to 100? or maybe difference of to squares, but i can't solve it.
 [/INST] 
Useful fact:
$$
\sqrt{a\pm \sqrt{b}}=
\sqrt{\frac{a+\sqrt{a^2-b}}{2}} \pm
\sqrt{\frac{a-\sqrt{a^2-b}}{2}} \tag{1}$$
Proof:
\begin{align}
  RHS^2 &=
  \frac{a+\sqrt{a^2-b}}{2}+\frac{a-\sqrt{a^2-b}}{2} \pm 2
  \sqrt{\left( \frac{a+\sqrt{a^2-b}}{2} \right)
        \left( \frac{a-\sqrt{a^2-b}}{2} \right)} \\
  &= a \pm 2\sqrt{\frac{a^2-(a^2-b)}{4}} \\
  &= a \pm 2\sqrt{\frac{b}{4}} \\
  &= a \pm \sqrt{b} \\
  &= LHS^2 \\
  LHS &= RHS \qquad (a^2 \ge b \ge 0)
\end{align}

Let
$$S_{\pm}=\sum_{j=1}^{n^2-1} \sqrt{n \pm \sqrt{j}} \tag{2}$$
and reverse the order of summation by taking $\, k=n^2-j$, then 
$$S_{\pm}=\sum_{k=1}^{n^2-1} \sqrt{n \pm \sqrt{n^2-k}} \tag{3}$$
Now by $(1)$, $(2)$ and $(3)$,
$$S_{\pm}=\frac{S_+ \pm S_-}{\sqrt{2}} \implies \frac{S_+}{S_-}=1+\sqrt{2}$$
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