We know that the Maclaurin series for $e^x$ is given by $\sum_{k=0}^{\infty} \frac{x^{k}}{k !}$. So, if we replace $x$ with $2x$ in this series, we get the Maclaurin series for $e^{2x}$, which is $\sum_{k=0}^{\infty} \frac{(2x)^{k}}{k !}$. The first four nonzero
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