As the series I got
$$ \sum_{n=0}^\infty (-1)^n(4x)^n/n! $$
which I think is right.
However, I am not sure how to get the first four non zero terms.
2026-04-05 18:40:09.1775414409
Taylor Series expansion and first four terms of $7x^2 e^{-4x}$
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We have
$$7x^2 e^{-4x}=7x^2\sum_{n=0}^\infty \frac{(-4x)^n}{n!}=7\sum_{n=0}^\infty\frac{(-4)^n}{n!}x^{n+2}=7\sum_{n=2}^\infty\frac{(-4)^{n-2}}{(n-2)!}x^{n}$$