Chapter 11 - Section 11.10 - Taylor and Maclaurin Series - 11.10 Exercises - Page 808: 42

$=\Sigma_{n=0}^{\infty}\frac{(3^{n}-2^{n})x^{n}}{n!}$

$f(x)=e^{3x}-e^{2x}$ Since, $e^{x}=\Sigma_{n=0}^{\infty}\frac{x^{n}}{n!}$ $e^{3x}-e^{2x}=\Sigma_{n=0}^{\infty}\frac{(3x)^{n}}{n!}-\Sigma_{n=0}^{\infty}\frac{(2x)^{n}}{n!}$ $=\Sigma_{n=0}^{\infty}\frac{(3^{n}-2^{n})x^{n}}{n!}$