Answer
$1-5x+\dfrac{25}{2}x^2-\dfrac{125}{6}x^3+\dfrac{625}{24}x^4-...$
Work Step by Step
Since, we know that the Maclaurin Series for $e^x$ is defined as:
$e^x=\Sigma_{n=0}^\infty \dfrac{x^n}{n!}=1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+...$
Now, $e^{-5x}=\Sigma_{n=0}^\infty \dfrac{(-5x)^n}{n!}=1-5x+\dfrac{(-5x)^2}{2!}+\dfrac{(-5x)^3}{3!}-\dfrac{(-5x)^4}{4!}...$
or, $=1-5x+\dfrac{25}{2}x^2-\dfrac{125}{6}x^3+\dfrac{625}{24}x^4-...$
