6.65. Expand the function $f(x)=e^{3 x}$ using the Maclaurin series formula.
Solution. We will use the expansion table, setting $u=3 x$:

$$
e^{u}=1+\frac{u}{1!}+\frac{u^{2}}{2!}+\ldots+\frac{u^{n}}{n!}+\ldots=1+\frac{3 x}{1}+\frac{9 x^{2}}{2}+\ldots+\frac{3^{n} x^{n}}{n!}+\ldots
$$