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Notice that the sequence of differences is constant. We know how to find the generating function for any constant sequence. So denote the generating function for \(1, 3, 5, 7, 9, \ldots\) by \(A\text{.}\) We have
\begin{align*}
A \amp = 1 + 3x + 5x^2 + 7x^3 + 9x^4 + \cdots\\
\underline{-\qquad xA} \amp \underline{\,\,= 0 + x + 3x^2 + 5x^3 + 7x^4 + 9x^5 + \cdots}\\
(1-x)A \amp = 1 + 2x + 2x^2 + 2x^3 + 2x^4 + \cdots
\end{align*}
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