Question

Найдите сумму: 1/1+√2 + 1/√2+√3 + . . . + 1/√2001+√2002

Answer 1

$\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{2001}+\sqrt{2002}}=\dfrac{1-\sqrt{2}}{(1+\sqrt{2})(1-\sqrt{2})}+\\ \\ \\ +\dfrac{\sqrt{2}-\sqrt{3}}{(\sqrt{2}+\sqrt{3})(\sqrt{2}-\sqrt{3})}+...+\dfrac{\sqrt{2001}-\sqrt{2002}}{(\sqrt{2001}+\sqrt{2002})(\sqrt{2001}-\sqrt{2002})}=\\ \\ \\ =\dfrac{1-\sqrt{2}}{1-2}+\dfrac{\sqrt{2}-\sqrt{3}}{2-3}+...+\dfrac{\sqrt{2001}-\sqrt{2002}}{2001-2002}=-1+\sqrt{2}-\sqrt{2}+\sqrt{3}-\\ \\ \\ -...-\sqrt{2001}+\sqrt{2002}=-1+\sqrt{2002}$