Respuesta :

dalendrk
[tex]\dfrac{2}{3}-\dfrac{4}{9}+\dfrac{8}{27}-\dfrac{16}{81}+...\\\\a_1=\dfrac{2}{3};\ a_2=-\dfrac{4}{9}\\\\r=a_2:a_1\to r=-\dfrac{4}{9}:\dfrac{2}{3}=-\dfrac{4}{9}\cdot\dfrac{3}{2}=-\dfrac{2}{3}\\\\|r|=\left|-\dfrac{2}{3}\right|=\dfrac{2}{3} \ \textless \ 1\\\\therefore\\\\S=\dfrac{a_1}{1-r}\\\\subtitute\\\\S=\dfrac{\frac{2}{3}}{1-\left(-\frac{2}{3}\right)}=\dfrac{\frac{2}{3}}{1+\frac{2}{3}}=\dfrac{\frac{2}{3}}{\frac{5}{3}}=\dfrac{2}{3}\cdot\dfrac{3}{5}=\boxed{\frac{2}{5}}[/tex]

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