Answer
$1-3x^2+\frac{9}{2}x^4-\frac{9}{2}x^6+...+\frac{1}{n!}(-3x^2)^n+...$
Work Step by Step
We are given $f(x)=e^{-3x^2}$
for $r$ in $(-\infty,\infty)$
The Taylor series for $e^x$ is
$1+x+\frac{1}{2!}x^2+\frac{1}{3!}x^3+...+\frac{1}{n!}x^n+...$
The Taylor series for $f(x)=e^{-3x^2}$ is
$f(x)=1+(-3x^2)+\frac{1}{2!}(-3x^2)^2+\frac{1}{3!}(-3x^2)^3+...+\frac{1}{n!}(-3x^2)^n+...$
$=1-3x^2+\frac{9}{2}x^4-\frac{9}{2}x^6+...+\frac{1}{n!}(-3x^2)^n+...$
