Ibrahim Rosales

2022-07-23

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 e2x and then multiply it by $({x}^{2}+4)$?

Reinfarktq6

Beginner2022-07-24Added 18 answers

$({x}^{2}+4){e}^{2x}=({x}^{2}+4)\sum _{n\ge 0}\frac{(2x{)}^{n}}{n!}=\sum _{n\ge 2}{x}^{n}\cdot \frac{{2}^{n-2}}{(n-2)!}+\sum _{n\ge 0}{x}^{n}\cdot \frac{{2}^{n}\cdot 4}{n!}=\phantom{\rule{0ex}{0ex}}=\sum _{n\ge 0}{x}^{n}\cdot \left(\frac{n(n-1){2}^{n-2}+{2}^{n+2}}{n!}\right)\phantom{\rule{thinmathspace}{0ex}}.$

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