[tex3]\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{n}+\sqrt{n+1}}=10[/tex3]
Resposta
n=120
Moderador: [ Moderadores TTB ]
Valeu chefe....AHAHAHAHAH bem burro mais bem forteMafIl10 escreveu: ↑Qui 29 Mar, 2018 12:27Fala frango kkkk
[tex3]\frac{1}{\sqrt{n}+\sqrt{n+1}}=\frac{\sqrt{n}-\sqrt{n+1}}{\sqrt{n}^{2}-(\sqrt{n+1})^{2}}=\sqrt{n+1}-\sqrt{n}\\
\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+\sqrt{5}-\sqrt{4}+....+\sqrt{n+1}-\sqrt{n}=11\\
-1+\sqrt{n+1}=11\\
n+1=121\\
n=120[/tex3]
Não seria [tex3]\sqrt{n+1}-1= 10 \rightarrow \sqrt{n+1} = 11 \rightarrow n=120 ?MatheusBorges escreveu: ↑Qui 29 Mar, 2018 12:27Fala frango kkkk
[tex3]\frac{1}{\sqrt{n}+\sqrt{n+1}}=\frac{\sqrt{n}-\sqrt{n+1}}{\sqrt{n}^{2}-(\sqrt{n+1})^{2}}=\sqrt{n+1}-\sqrt{n}\\
\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+\sqrt{5}-\sqrt{4}+....+\sqrt{n+1}-\sqrt{n}=11\\
-1+\sqrt{n+1}=11\\
n+1=121\\
n=120[/tex3]