Estude! Com Prof. Caju

Problema Proposto
43 - Dado os pontos consecutivos A. B, C,
D, E, ..., de modo que :
AB = [tex3]\frac{1}{2}[/tex3] , BC= [tex3]\frac{1}{3}[/tex3], CD=[tex3]\frac{1}{4}[/tex3], DE=[tex3]\frac{1}{9}[/tex3] , EF = [tex3]\frac{1}{8}[/tex3]...
Calcular a soma limite de x onde x = AB + BC + CD + DE + ........ .

[tex3]\mathsf{
x = (\underbrace{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...}_{PG:q=\frac{1}{2}})+(\underbrace{\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...}_{PG:q=\frac{1}{3}})=\frac{\frac{1}{2}}{1-\frac{1}{2}}+\frac{\frac{1}{3}}{1 - \frac{1}{3}}=\\
x = 1 + \frac{1}{2} \therefore \boxed{\color{red}x = \frac{3}{2}=1,5}

}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\mathsf{
x = (\underbrace{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...}_{PG:q=\frac{1}{2}})+(\underbrace{\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...}_{PG:q=\frac{1}{3}})=\frac{\frac{1}{2}}{1-\frac{1}{2}}+\frac{\frac{1}{3}}{1 - \frac{1}{3}}=\\
x = 1 + \frac{1}{2} \therefore \boxed{\color{red}x = \frac{3}{2}=1,5}

}[/tex3]