Chapter 11 - Section 11.10 - Taylor and Maclaurin Series - 11.10 Exercises - Page 771: 40

$\Sigma_{n=1}^{\infty}(-1)^{n-1}\frac{x^{(3n+2)}}{n}$

$x^{2} \cdot ln(1+x^{3})=x^{2} \Sigma_{n=1}^{\infty}(-1)^{n-1}\frac{(x^{3})^{n})^{n}}{n}$ $=x^{2} \Sigma_{n=1}^{\infty}(-1)^{n-1}\frac{x^{3n}}{n}$ $=\Sigma_{n=1}^{\infty}(-1)^{n-1}\frac{x^{(3n+2)}}{n}$