Again we call the generating function for the sequence \(A\text{.}\) Using differencing:
\begin{align*}
A \amp = 1 + 4x + 9x^2 + 16x^3 + \cdots\\
\underline{-\qquad xA} \amp \underline{\,\, = 0 + x + 4x^2 + 9x^3 + 16x^4 + \cdots}\\
(1-x)A \amp = 1 + 3x + 5x^2 + 7x^3 + \cdots
\end{align*}
Since \(1 + 3x + 5x^2 + 7x^3 + \cdots = \d\frac{1+x}{(1-x)^2}\) we have \(A = \d\frac{1+x}{(1-x)^3}\text{.}\)