Question

Calcule a potência de (0,181818...)^2

Answer 1

Temos o seguinte:

$0,181818... = \frac{18}{100} + \frac{18}{10000} + \frac{18}{1000000} + \cdots = \sum_{n=1}^{\infty} \dfrac{18}{100^n}$

Calculando a soma infinita, com razão $q = \frac{1}{100}$:

$S_{\infty} = \dfrac{a_{1}}{1 - q} = \frac{ \frac{18}{100} }{1 - \frac{1}{100} } = \dfrac{18}{100} \cdot \dfrac{100}{99} = \dfrac{18}{99} = \dfrac{2}{11}$

Logo temos:

$0,181818... = \sum_{n=1}^{\infty} \dfrac{18}{100^n} = \dfrac{2}{11}$

Assim:

$(0,181818...)^2 = \left(\dfrac{2}{11} \right)^2 = \boxed{ \dfrac{4}{121} }$