The given problem asks for the first three terms of the power series:

$\sum_{n=0}^{\infty} \frac{(3)^n}{n!} x^{n+4}$

To find the first three terms, we substitute $n = 0$, $n = 1$, and $n = 2$ into the general formula.

Step-by-step calculation:

1. For $n = 0$: $\frac{(3)^0}{0!} x^{0+4} = \frac{1}{1} x^4 = x^4$

2. For $n = 1$: $\frac{(3)^1}{1!} x^{1+4} = \frac{3}{1} x^5 = 3x^5$

3. For $n = 2$: $\frac{(3)^2}{2!} x^{2+4} = \frac{9}{2} x^6 = \frac{9}{2}x^6$

Therefore, the first three terms of the power series are:

$x^4 + 3x^5 + \frac{9}{2}x^6$

Let me know if you want further clarification or more details!

Here are 5 related questions to deepen your understanding:

1. How do we determine the general term of a power series?
2. What is the radius of convergence for this particular power series?
3. How would the series change if the coefficient $3^n$ were replaced with $n^3$?
4. How do factorial terms like $n!$ affect the growth of a power series?
5. How does the exponent of $x$ impact the behavior of the power series for large values of $x$?

Tip: When working with series, always remember to check for convergence using appropriate tests, such as the ratio test, if you're dealing with an infinite number of terms.