Answer
$x^4+2x^5+2x^6+\frac{4x^7}{3}+...+\frac{(2x)^{n+4}}{n!}+...$
Work Step by Step
We are given $f(x)=x^4e^{2x}$
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 $f(x)=x^4e^{2x}$ is
$f(x)=x^4.1+x^4(2x)+x^4\frac{1}{2!}(2x)^2+x^4\frac{1}{3!}(2x)^3+...+x^4\frac{1}{n!}(2x)^n+...$
$=x^4+2x^5+2x^6+\frac{4x^7}{3}+...+\frac{(2x)^{n+4}}{n!}+...$
