# Find the Maclaurin series for f(x)=(x^2+4)e^(2x) and use it to calculate the 1000th derivative of f(x) at x=0. Is it possible to just find the Maclaurin series for e^(2x) and then multiply it by (x^2+4)?

Find the Maclaurin series for $f\left(x\right)=\left({x}^{2}+4\right){e}^{2x}$ and use it to calculate the 1000th derivative of $f\left(x\right)$ at $x=0$. Is it possible to just find the Maclaurin series for e2x and then multiply it by $\left({x}^{2}+4\right)$?
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Reinfarktq6
$\left({x}^{2}+4\right){e}^{2x}=\left({x}^{2}+4\right)\sum _{n\ge 0}\frac{\left(2x{\right)}^{n}}{n!}=\sum _{n\ge 2}{x}^{n}\cdot \frac{{2}^{n-2}}{\left(n-2\right)!}+\sum _{n\ge 0}{x}^{n}\cdot \frac{{2}^{n}\cdot 4}{n!}=\phantom{\rule{0ex}{0ex}}=\sum _{n\ge 0}{x}^{n}\cdot \left(\frac{n\left(n-1\right){2}^{n-2}+{2}^{n+2}}{n!}\right)\phantom{\rule{thinmathspace}{0ex}}.$