Answer
$\Sigma_{n=0}^{\infty} (-1)^n \dfrac{3^n}{n !}x^n$
Work Step by Step
We know that the Taylor series for $e^x=\Sigma_{n=0}^{\infty} \dfrac{x^n}{n!}=1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+.....$
Replace $x$ with $-3x$.
Thus, our required series is:
$e^{-3x}=1-3x+\dfrac{9x^2}{2}-\dfrac{27 x^3}{3!}+....+\dfrac{x^n}{n !}\\=\Sigma_{n=0}^{\infty} (-1)^n \dfrac{3^n}{n !}x^n$
