Evaluate the sum \(\sum\limits^{\infty}_{k=1}(2k+1)x^2k\) for \(\lvert x \rvert < 1\).

\((2k+1)x^2k\) looks familiar in that it is the derivative of \(x^{2k+1}\). Let’s work backwards to derive a solution.

\[\begin{split} \sum\limits^{\infty}_{k=1}x^k &= \frac{x}{1-x} \\ \sum\limits^{\infty}_{k=1}(x^2)^k &= \frac{x^2}{1-x^2} \\ \sum\limits^{\infty}_{k=1}xx^{2k} &= \frac{x^3}{1-x^2} \\ \sum\limits^{\infty}_{k=1}x^{2k+1} &= \frac{x^3}{1-x^2} \\ \sum\limits^{\infty}_{k=1}(2k+1)x^{2k} &= \frac{3x^2(1-x^2)+2x^4}{(1-x^2)^2} \\ \sum\limits^{\infty}_{k=1}(2k+1)x^{2k} &= \frac{3x^2-x^4}{(1-x^2)^2} \\ \end{split}\]

for \(\lvert x \rvert < 1\).