Series of Nested Radicals

Question

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.

Answer 1

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 $(3)$,

$$S_{\pm}=\frac{S_+ \pm S_-}{\sqrt{2}} \implies \frac{S_+}{S_-}=1+\sqrt{2}$$