Science:Math Exam Resources/Courses/MATH101/April 2014/Question 01 (i)/Solution 2

Since the Maclaurin series for $e^x$

$\begin{array}{r}{e}^x=\sum _{n=0}^{\mathrm{\infty }}\frac{x^n}{n!},\end{array}$

by plugging $3x$ in the position of $x$, we have the Maclaurin series for $e^{3x}$:

$\begin{array}{r}{e}^{3x}=\sum _{n=0}^{\mathrm{\infty }}\frac{3^n{x}^n}{n!}.\end{array}$

Since the coefficient of the fifth degree term in this Maclaurin series is $\frac{3^5}{5!}$, the answer is $\frac{3^5}{5!}=\frac{243}{120}$.