Answer
$2-\frac{2x^2}{3}+\frac{2x^4}{9}-\frac{2x^6}{27}+...+\frac{2(-x^{2+n})}{3^n}$
Work Step by Step
We are given $f(x)=\frac{6}{3+x^2}=\frac{2}{1-(-\frac{x^2}{3})}$
for $r$ in $(-\infty,\infty)$
The Taylor series for $\frac{1}{1-x}$ is
$1+x+x^2+x^3+...+x^n+...$
The Taylor series for $f(x)=\frac{2}{1-(-\frac{x^2}{3})}$ is
$f(x)=2.1+2(-\frac{x^2}{3})+2(-\frac{x^2}{3})^2+2(-\frac{x^2}{3})^3...+2(-\frac{x^2}{3})^n+...$
$=2-\frac{2x^2}{3}+\frac{2x^4}{9}-\frac{2x^6}{27}+...+\frac{2(-x^{2+n})}{3^n}$