jmc

Rationalizing the denominator, we get $$\frac{2}{\sqrt{n}+\sqrt{n+2}}=\frac{2(\sqrt{n+2}-\sqrt{n})}{(\sqrt{n+2}+\sqrt{n})(\sqrt{n+2}-\sqrt{n})}=\frac{2(\sqrt{n+2}-\sqrt{n})}{(n+2)-n}=\sqrt{n+2}-\sqrt{n}.$$Therefore, $$\begin{array}{rl}\sum _{n=1}^{99}\frac{2}{\sqrt{n}+\sqrt{n+2}}& =\sum _{n=1}^{99}(\sqrt{n+2}-\sqrt{n})\\ & =(\sqrt{3}-1)+(\sqrt{4}-\sqrt{2})+(\sqrt{5}-\sqrt{3})+\cdots +(\sqrt{100}-\sqrt{98})+(\sqrt{101}-\sqrt{99})\\ & =\sqrt{100}+\sqrt{101}-1-\sqrt{2}\\ & =\boxed{\sqrt{101}-\sqrt{2}+9}.\end{array}$$