<s>[INST] 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. 
 [/INST] We can rewrite the sum as:
$$\sum_{n=1}^{99} \frac{2}{\sqrt{n} + \sqrt{n+2}}\frac{\sqrt{n+2} - \sqrt{n}}{\sqrt{n+2} - \sqrt{n}}=
\sum_{n=1}^{99} \sqrt{n+2} - \sqrt{n}$$
It's a telescopic sum, where most of the terms are cancelled.
We can expand it:
$$\sum_{n=1}^{99} \sqrt{n+2} - \sqrt{n}=
(\sqrt{3}-\sqrt{1})+(\sqrt{4}-\sqrt{2})+(\sqrt{5}-\sqrt{3})+\dots+(\sqrt{101}-\sqrt{99})=\sqrt{100}+\sqrt{101}-\sqrt{1}-\sqrt{2}$$
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