Question

Vă rog !! e doar un exercițiu! !!

Answer 1

$S=\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{2017}+\sqrt{2018}}$

$S=\frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{3}+\sqrt{2}}+...+\frac{1}{\sqrt{2018}+\sqrt{2017}}$

$S=\frac{\sqrt{2}-1}{(\sqrt{2}-1)*(\sqrt{2}+1)}+\frac{\sqrt{3}-\sqrt{2}}{(\sqrt{3}+\sqrt{2})*(\sqrt{3}-\sqrt{2})}+...+\frac{\sqrt{2018}-\sqrt{2017}}{(\sqrt{2018}+\sqrt{2017})*(\sqrt{2018}-\sqrt{2017})}$

$S=\frac{\sqrt{2}-1}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+...+\frac{\sqrt{2018}-\sqrt{2017}}{2018-2017}$

$S=\frac{\sqrt{2}-1}{1}+\frac{\sqrt{3}-\sqrt{2}}{1}+...+\frac{\sqrt{2018}-\sqrt{2017}}{1}$

$S=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{2018}-\sqrt{2017}$

$S=-1+\sqrt{2}-\sqrt{2}+\sqrt{3}-\sqrt{3}+...+\sqrt{2017}-\sqrt{2017}+\sqrt{2018}$

$S=-1+\sqrt{2018}$

$\boxed{S=\sqrt{2018}-1}$