What is \frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\cdot\cdot\cdot+\frac{1}{\sqrt{2020}+\sqrt{2021}} equal to?

Explanation

Rationalizing the general term \frac{1}{\sqrt{n}+\sqrt{n+1}} yields \sqrt{n+1} - \sqrt{n}. The sum becomes a telescoping series: (\sqrt{2}-1) + (\sqrt{3}-\sqrt{2}) + \dots + (\sqrt{2021}-\sqrt{2020}), leaving \sqrt{2021} - 1.

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