The sequence is given by the formula:

$a_n = 4(2)^n$

Where $n$ represents the position of a term in the sequence.

We will calculate the first three terms of the sequence by substituting $n = 1, 2, 3$ into the formula.

For $n = 1$:

$a_1 = 4(2)^1 = 4 \times 2 = 8$

For $n = 2$:

$a_2 = 4(2)^2 = 4 \times 4 = 16$

For $n = 3$:

$a_3 = 4(2)^3 = 4 \times 8 = 32$

Thus, the first three terms of the sequence are:

$8, 16, 32$

Would you like further clarification or details about how I arrived at these terms? Here are a few related questions to explore:

1. How would the sequence change if the formula was $a_n = 3(2)^n$?
2. What is the general behavior of exponential sequences like this one?
3. How can you prove that $a_n = 4(2)^n$ grows exponentially?
4. What is the sum of the first five terms of this sequence?
5. How would the sequence behave if you had a negative exponent (e.g., $a_n = 4(2)^{-n}$)?

Tip: When dealing with exponential sequences, always pay attention to the base of the exponent. The base 2 in this case makes the sequence grow quickly as $n$ increases.