Let's analyze the sequence formula given in the image:

The sequence is defined as: $b_j = \frac{(-1)^j}{3^j} \quad \text{for } j \geq 0$

We are tasked to find the first four terms of the sequence, corresponding to $j = 0, 1, 2, 3$.

Step-by-step calculation:

1. For $j = 0$: $b_0 = \frac{(-1)^0}{3^0} = \frac{1}{1} = 1$

2. For $j = 1$: $b_1 = \frac{(-1)^1}{3^1} = \frac{-1}{3} = -\frac{1}{3}$

3. For $j = 2$: $b_2 = \frac{(-1)^2}{3^2} = \frac{1}{9}$

4. For $j = 3$: $b_3 = \frac{(-1)^3}{3^3} = \frac{-1}{27}$

First four terms:

$b_0 = 1, \quad b_1 = -\frac{1}{3}, \quad b_2 = \frac{1}{9}, \quad b_3 = -\frac{1}{27}$

Would you like further clarification or a deeper explanation of this sequence? Here are five questions to expand the topic:

1. What is the general behavior of this sequence as $j$ becomes large?
2. Can you find the sum of the infinite series generated by this sequence?
3. How does this sequence relate to alternating geometric series?
4. What is the practical significance of sequences like this in mathematics?
5. How can this sequence be represented in sigma notation for summation?

Tip: When solving sequences with powers, always double-check whether the exponent affects the sign or magnitude!