Question

Considerando a soma infinita: x+2x/3+4x/9+ ... = 12, determine o valor de x


a.
3


b.
8


c.
6


d.
2


e.
4

Answer 1

$\boxed{\begin{array}{l}\sf x+\dfrac{2x}{3}+\dfrac{4x}{9}+... =12\\\sf x+\dfrac{2x}{3}+\dfrac{4x}{9}+\dfrac{8x}{27}+...=12\\\sf x+x\cdot\bigg(\dfrac{2}{3}+\dfrac{4}{9}+\dfrac{8}{27}+...\bigg)=12\\\sf a_1=\dfrac{2}{3}\\\sf a_{2}=\dfrac{4}{9}\\\sf q=\dfrac{\frac{4}{9}}{\frac{2}{3}}\\\sf q=\dfrac{\diagdown\!\!\!\!4^2}{\diagup\!\!\!\!9_3}\cdot\dfrac{\diagup\!\!\!\!3}{\diagdown\!\!\!2}\\\sf q=\dfrac{2}{3}\\\sf S_n=\dfrac{a_1}{1-q} \\\sf S_n=\dfrac{\frac{2}{3}}{1-\frac{2}{3}}\\\sf S_n=\dfrac{\frac{2}{3}}{\frac{1}{3}}\\\sf S_n=\dfrac{2}{3}\cdot3=2\\\sf x+2x=12\\\sf 3x=12\\\sf x=\dfrac{12}{3}\\\sf x=4\end{array}}$