Question

$\frac{1}{ \sqrt{2 + 1} } + \frac{1}{ \sqrt{3} + \sqrt{2} } + \frac{1}{ \sqrt{4 } + \sqrt{3} } + ... + \frac{1}{ \sqrt{2020+ \sqrt{2019} } }$
​Пожалуйста, решите!

Answer 1

$\dfrac{1}{\sqrt{2}+\sqrt{1}}+\dfrac{1}{\sqrt{3}+\sqrt{2}}+\dfrac{1}{\sqrt{4}+\sqrt{3}}+...+\dfrac{1}{\sqrt{2020}+\sqrt{2019}}=\\ \\ \\ =\dfrac{\sqrt{2}-\sqrt{1}}{(\sqrt{2}+\sqrt{1})(\sqrt{2}-\sqrt{1})}+\dfrac{\sqrt{3}-\sqrt{2}}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}+\dfrac{\sqrt{4}-\sqrt{3}}{(\sqrt{4}+\sqrt{3})(\sqrt{4}-\sqrt{3})}+\\ \\ \\ +...+\dfrac{\sqrt{2020}-\sqrt{2019}}{(\sqrt{2020}+\sqrt{2019})(\sqrt{2020}-\sqrt{2019})}=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+\sqrt{4}-$

$-\sqrt{3}+...+\sqrt{2020}-\sqrt{2019}=-1+\sqrt{2020}$