Resposta :
[tex]P.G(1/3,2/9,4/27,...)[/tex]
[tex]a_{1} = 1/3[/tex]
[tex]a_{2}=2/9[/tex]
[tex]q = a_{2}/a_{1} => (2/9)/(1/3) => (2/9)*(3/1) => (2/3)*(1/1) => q = 2/3[/tex]
[tex]0 < q < 1[/tex]: Soma infinita
[tex]S_{n} = a_{1} / (1 - q)[/tex]
[tex]S_{n} = (1/3) / (1 - [2/3])[/tex]
[tex]S_{n} = (1/3)/ ([3/3]-[2/3])[/tex]
[tex]S_{n} = (1/3)/([3-2]/3)[/tex]
[tex]S_{n} = (1/3)/(1/3)[/tex]
[tex]S_{n}=1[/tex]
[tex]a_{1} = 1/3[/tex]
[tex]a_{2}=2/9[/tex]
[tex]q = a_{2}/a_{1} => (2/9)/(1/3) => (2/9)*(3/1) => (2/3)*(1/1) => q = 2/3[/tex]
[tex]0 < q < 1[/tex]: Soma infinita
[tex]S_{n} = a_{1} / (1 - q)[/tex]
[tex]S_{n} = (1/3) / (1 - [2/3])[/tex]
[tex]S_{n} = (1/3)/ ([3/3]-[2/3])[/tex]
[tex]S_{n} = (1/3)/([3-2]/3)[/tex]
[tex]S_{n} = (1/3)/(1/3)[/tex]
[tex]S_{n}=1[/tex]
a1 =1/3 = 3^-1
q = a2 = 2/9 ==> 2 . 3 ==> 2
a1 1/3 9 1 3
n= ?
an =?
Para determinarmos a soma precisamos saber qual o termo a ser calculado ou dando o último termos.