Question

$\frac{1}{1+ \sqrt{2} }+ \frac{1}\sqrt{2}+\sqrt{3} }+\frac{1}\sqrt{3}+\sqrt{4} }+...\frac{1}\sqrt{99}+\sqrt{100} }$

Answer 1

$\dfrac{1}{1+\sqrt2}+\dfrac{1}{\sqrt2+\sqrt3}+\dfrac{1}{\sqrt3+\sqrt4}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}=\\ \\ \\ =\dfrac{1}{\sqrt2+1}+\dfrac{1}{\sqrt3+\sqrt2}+\dfrac{1}{\sqrt4+\sqrt3}+...+\dfrac{1}{\sqrt{100}+\sqrt{99}}= \\ \\ ( amplificam cu numitorul conjugat la fiecare fractie:)$

$=\dfrac{\sqrt2-1}{(\sqrt2-1)(\sqrt2+1)}+\dfrac{\sqrt3-\sqrt2}{(\sqrt3-\sqrt2)(\sqrt3+\sqrt2)}+\\ \\ + \dfrac{\sqrt4-\sqrt3}{(\sqrt4-\sqrt3)(\sqrt4+\sqrt3)}+...+\dfrac{\sqrt{100}-\sqrt{99}}{(\sqrt{100}-\sqrt{99})(\sqrt{100}+\sqrt{99})}= \\ \\ \\ =\dfrac{\sqrt2 -1}{2-1}+ \dfrac{\sqrt3-\sqrt2}{3-2}+\dfrac{\sqrt4-\sqrt3}{4-3}+...+\dfrac{\sqrt{100}-\sqrt{99}}{100-99} = \\ \\ =\sqrt2-1+\sqrt3-\sqrt2+\sqrt4-\sqrt3+...+\sqrt{100}-\sqrt{99} = \\ \\ =-1+\sqrt{100} = \sqrt{100}-1$