Question

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

Answer 1

Răspuns:

Explicație pas cu pas:

Amplificam cu conjugatele numitorilor.

Poza!

Answer 2

$S = \dfrac{1}{1+\sqrt 2}+\dfrac{1}{\sqrt 2+\sqrt 3}+...+\dfrac{1}{\sqrt{2017}+\sqrt{2018}} \\ \\ S = \sum\limits_{n=1}^{2017}\dfrac{1}{\sqrt n+\sqrt {n+1}} = \sum\limits_{n=1}^{2017}\dfrac{\sqrt n - \sqrt{n+1}}{(\sqrt n - \sqrt{n+1})(\sqrt n+\sqrt {n+1})} = \\ \\ =\sum\limits_{n=1}^{2017}\dfrac{\sqrt n - \sqrt{n+1}}{n-(n+1)} = \sum\limits_{n=1}^{2017}\dfrac{\sqrt n - \sqrt{n+1}}{-1} = \sum\limits_{n=1}^{2017}(\sqrt{n+1}-\sqrt n)=$

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