[Resolução] 10.13.18

[Aracy]

Enunciado:
18. \(\frac{x}{1+x-2x^2} = \frac{1}{3} \sum_{n=1}^{\infty} (1-(-2)^n)x^n \hspace{1.05cm} (|x| < \frac{1}{2})\)

Resposta:
Sabendo que (dica do exercício):
\(
\begin{eqnarray}
\frac{x}{1+x-2x^2} &=& \frac{1}{3} \left(\frac{3x}{1+x-2x^2}\right) \\
&=& \frac{1}{3} \left(\frac{1}{1-x} - \frac{1}{1+2x}\right)
\end{eqnarray}
\)

Relembrando as fórmulas para as séries geométricas...
\(
\begin{eqnarray}
\frac{1}{1-x} &=& \sum_{n=0}^{\infty} x^n \\
\frac{1}{1+2x} &=&\sum_{n=0}^{\infty} (-1)^n(2x)^n
\end{eqnarray}
\)

Dessa forma, temos:
\(
\begin{eqnarray}
\frac{x}{1+x-2x^2} &=& \frac{1}{3}\left(\frac{1}{1-x} - \frac{1}{1+2x}\right) \\
&=& \frac{1}{3} \left(\sum_{n=0}^{\infty} x^n - \sum_{n=0}^{\infty} (-1)^n(2x)^n \right)\\
&=&\frac{1}{3} \sum_{n=0}^{\infty} (1-(-2)^n)x^n
\end{eqnarray}
\)

[macarrao_carbonara]

Entendi as contas, fez bastante sentido mesmo, obrigado