Respuesta :

Answer:

[tex]\frac{2}{3}[/tex]

Step-by-step explanation:

The sum to infinity of a GP is

[tex]S_{infinity}[/tex] = [tex]\frac{a}{1-r}[/tex] ; | r | < 1

where a is the first term and r the common ratio

Here a = 1 and r = - [tex]\frac{1}{2}[/tex] ÷ 1 = - [tex]\frac{1}{2}[/tex] , thus

[tex]S_{infinity}[/tex] = [tex]\frac{1}{1-(-\frac{1}{2}) }[/tex] = [tex]\frac{1}{1+\frac{1}{2} }[/tex] = [tex]\frac{1}{\frac{3}{2} }[/tex] = [tex]\frac{2}{3}[/tex]

Answer:

2/3

Step-by-step explanation:

1-1/2+1/4.....

[tex]r=\frac{\frac{-1}{2} }{1} =-\frac{1}{2} \\if~|r|<1,then~s_{\infty}=\frac{a}{1-r} ,a=first~term.\\s_{\infty}=\frac{1}{1-(\frac{-1}{2}) } =\frac{1}{1+\frac{1}{2} } =\frac{1}{\frac{3}{2} } =\frac{2}{3}[/tex]

Otras preguntas