Evaluate $\sum_{n = 1}^{99} \frac{2}{\sqrt{n} + \sqrt{n + 2}}.$ Evaluate $\sum_{n = 1}^{99} \frac{2}{\sqrt{n} + \sqrt{n + 2}}.$
Listing out a few terms and simplifying the denominators gets me $$-1+\sqrt{3}+2-\sqrt{2}-\sqrt{3}+\sqrt{5}\cdots-\sqrt{97}+3\sqrt{11}-7\sqrt{2}+10-3\sqrt{11}+\sqrt{101}.$$
A lot of these terms cancel, and I think that the first and last terms will be left.
I feel like I'm close, but missing something. 

Let $ m \in\mathbb{N}^{*} : $
\begin{aligned}\sum_{n=1}^{m}{\frac{2}{\sqrt{n}+\sqrt{n+2}}}&=\sum_{n=1}^{m}{\left(\sqrt{n+2}-\sqrt{n}\right)}\\ &=\sum_{n=1}^{m}{\left(\sqrt{n+2}-\sqrt{n+1}\right)}+\sum_{n=1}^{m}{\left(\sqrt{n+1}-\sqrt{n}\right)}\\ \sum_{n=1}^{m}{\frac{2}{\sqrt{n}+\sqrt{n+2}}}&=\sqrt{m+2}-\sqrt{2}+\sqrt{m+1}-1\end{aligned}
