Exercice 14 --- (id : 1092)
Activités numériques II: Exercice 14
correction
1 $$\begin{align*} &\dfrac{1}{\sqrt{n+1}+\sqrt{n}}\\ =&\dfrac{\sqrt{n+1}-\sqrt{n}}{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}\\ =&\dfrac{\sqrt{n+1}-\sqrt{n}}{(\sqrt{n+1})^2-(\sqrt{n})^2}\\ =&\dfrac{\sqrt{n+1}-\sqrt{n}}{(n+1)-n}\\ =&\dfrac{\sqrt{n+1}-\sqrt{n}}{1}\\ =&\sqrt{n+1}-\sqrt{n} \end{align*}$$
2 $$\begin{align*} &\dfrac{1}{\sqrt{2}+1}+\dfrac{1}{\sqrt{3}+\sqrt{2}}+\dfrac{1}{2+\sqrt{3}}+...+\dfrac{1}{10+\sqrt{99}}\\ &\dfrac{1}{\sqrt{2}+\sqrt{1}}+\dfrac{1}{\sqrt{3}+\sqrt{2}}+\dfrac{1}{\sqrt{4}+\sqrt{3}}+...+\dfrac{1}{\sqrt{100}+\sqrt{99}}\\ &=\left({\sqrt{2}-\sqrt{1}}\right)+\left({\sqrt{3}-\sqrt{2}}\right)+\left({\sqrt{4}-\sqrt{3}}\right)+...+\left({\sqrt{100}-\sqrt{99}}\right)\\ &=\sqrt{100}-\sqrt{1}=10-1=9 \end{align*}$$