Find the Maclaurin series for the given function.
-$e ^ { 4 x }$
A) $\sum _ { n = 0 } ^ { \infty } \frac { 4 ^ { n } x ^ { n } } { n ! }$
B) $\sum _ { \mathrm { n } = 0 } ^ { \infty } \frac { ( - 1 ) ^ { \mathrm { n } } 4 ^ { \mathrm { n } } \mathrm { x } ^ { \mathrm { n } } } { \mathrm { n } ! }$
C) $\sum _ { n = 1 } ^ { \infty } \frac { ( - 1 ) ^ { n } 4 ^ { n } x ^ { n } } { n ! }$
D) $\sum _ { n = 1 } ^ { \infty } \frac { 4 ^ { n } x ^ { n } } { n ! }$

Question

Quizplus · Accepted Answer

The Maclaurin series for the function $e^x$ is $\sum_{n=0}^\infty \frac{x^n}{n!}$. Thus, for $e^{4x}$, we just need to replace $x$ with $4x$: 
$$e^{4x} = \sum_{n=0}^\infty \frac{(4x)^n}{n!} = \sum_{n=0}^\infty \frac{4^n x^n}{n!}$$
Therefore, the Maclaurin series for $e^{4x}$ is $\boxed{	extbf{(A)}\ \sum _ { n = 0 } ^ { \infty } \frac { 4 ^ { n } x ^ { n } } { n ! }}$.

Find the Maclaurin Series for the Given Function e4xe ^ { 4 x }e4x

Find the Maclaurin Series for the Given Function $e ^ { 4 x }$