Question

find the maclaurin series for the given function
$e^{5x}$
a. $sum_{n = 1}^{infty}\frac{(-1)^n5^nx^n}{n!}$
b. $sum_{n = 1}^{infty}\frac{5^nx^n}{n!}$
c. $sum_{n = 0}^{infty}\frac{5^nx^n}{n!}$
d. $sum_{n = 0}^{infty}\frac{(-1)^n5^nx^n}{n!}$

Explanation:

Step1: Recall the Maclaurin series of \(e^t\)

The Maclaurin series of \(e^t=\sum_{n = 0}^{\infty}\frac{t^n}{n!}\), where \(n!=n\times(n - 1)\times\cdots\times1\) and \(0!=1\).

Step2: Substitute \(t = 5x\)

When we substitute \(t = 5x\) into the Maclaurin series of \(e^t\), we get \(e^{5x}=\sum_{n=0}^{\infty}\frac{(5x)^n}{n!}=\sum_{n = 0}^{\infty}\frac{5^n x^n}{n!}\)

Answer:

C. \(\sum_{n = 0}^{\infty}\frac{5^n x^n}{n!}\)