Ensino Superior ⇒ Simplifique Tópico resolvido

[guilmarangon]

[tex3]\frac{1}{\sqrt{2}+1} + \frac{1}{\sqrt{3}+\sqrt{2}} + \frac{1}{\sqrt{4}+\sqrt{3}}[/tex3]+......+[tex3]\frac{1}{\sqrt{100}+\sqrt{99}}[/tex3]

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[tex3]\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{99}+\sqrt{100}}=\sum_{n=1}^{99}\frac{1}{\sqrt{n}+\sqrt{n+1}}[/tex3]

[tex3]\frac{1}{\sqrt{n}+\sqrt{n+1}}=\frac{1}{\sqrt{n}+\sqrt{n+1}}.\frac{\sqrt{n}-\sqrt{n+1}}{\sqrt{n}-\sqrt{n+1}}=\frac{\sqrt{n}-\sqrt{n+1}}{(n)-(n+1)}=\sqrt{n+1}-\sqrt{n}[/tex3]

[tex3]\therefore \sum_{n=1}^{99}\frac{1}{\sqrt{n}+\sqrt{n+1}}=[/tex3]
[tex3]\sum_{n=1}^{99}\sqrt{n+1}-\sqrt{n}=(\cancel{\sqrt{2}}-\sqrt{1})+(\cancel{\sqrt{3}}-\cancel{\sqrt{2}})+(\cancel{\sqrt{4}}-\cancel{\sqrt{3}})+...+(\cancel{\sqrt{99}}-\cancel{\sqrt{98}})+(\sqrt{100}-\cancel{\sqrt{99}})=-\sqrt{1}+\sqrt{100}=-1+10=\boxed{9}[/tex3]

Qualquer dúvida me avise!

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Eu tinha errado um negocinho, mas já corrigi!