The sequence given is defined by the formula:

$a_n = 4(2)^n$

We are asked to find the first five terms of the sequence, starting with $n = 1$.

Let's calculate the terms one by one:

1. For $n = 1$: $a_1 = 4(2)^1 = 4 \times 2 = 8$

2. For $n = 2$: $a_2 = 4(2)^2 = 4 \times 4 = 16$

3. For $n = 3$: $a_3 = 4(2)^3 = 4 \times 8 = 32$

4. For $n = 4$: $a_4 = 4(2)^4 = 4 \times 16 = 64$

5. For $n = 5$: $a_5 = 4(2)^5 = 4 \times 32 = 128$

So, the first five terms of the sequence are:

$8, 16, 32, 64, 128$

Would you like further clarification or additional details?

Here are 5 related questions for further exploration:

1. How would the sequence change if the base of the exponent (2) was replaced by a different number?
2. Can you express the $n$-th term of this sequence using logarithms?
3. What is the common ratio in this sequence, and how does it help identify the sequence as geometric?
4. If the starting term was different (e.g., 6 instead of 4), how would the sequence be affected?
5. What happens to the sequence as $n$ tends to infinity?

Tip: In geometric sequences, each term is the product of the previous term and a constant (the common ratio). Understanding this property helps in solving more complex sequence problems.