The sequence provided is defined as:

$a_n = 2(2^n)$

We are tasked with finding the first five terms, starting with $n = 1$.

Let's calculate the terms:

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

2. For $n = 2$: $a_2 = 2(2^2) = 2 \times 4 = 8$

3. For $n = 3$: $a_3 = 2(2^3) = 2 \times 8 = 16$

4. For $n = 4$: $a_4 = 2(2^4) = 2 \times 16 = 32$

5. For $n = 5$: $a_5 = 2(2^5) = 2 \times 32 = 64$

Thus, the first five terms of the sequence are:

$4, 8, 16, 32, 64$

Let me know if you'd like more details or have any questions!

Related questions:

1. How would the sequence change if the base of the exponent were 3 instead of 2?
2. What is the general formula for the sum of the first $n$ terms of this sequence?
3. How does this sequence relate to geometric progressions?
4. What happens if we change the multiplier from 2 to another constant?
5. How do we derive the $n$-th term for this type of exponential sequence?

Tip: In exponential sequences, each term grows rapidly due to the power function. Understanding how the exponent affects growth is key to analyzing such sequences.