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When dealing with recursively defined sequences, understanding sequence terms is essential. A sequence is simply a list of numbers arranged in a specific order. In this case, the terms are generated based on a rule or pattern established previously.
The sequence is initiated with a starting value, in this case, the first term:

• The first term is given as \(a_0 = -4\).

By using the recurrence relation provided, each subsequent term is derived from the one preceding it. For the first four terms:

• The second term \(a_1\) is calculated as: \(-4 + 2 = -2\).
• The third term \(a_2\) is: \(-2 + 2 = 0\).
• Finally, the fourth term \(a_3\) becomes: \(0 + 2 = 2\).

Thus, the first four terms of the sequence are \(-4, -2, 0, 2\). By understanding these initial terms, you set a solid foundation for grasping the entire sequence.

A recurrence relation is a functional equation that defines each term of a sequence using the previous term(s). This is a crucial concept for understanding recursively defined sequences, as it offers the backbone for generating the sequence.
In this particular exercise, the recurrence relation given is: \(a_n = a_{n-1} + 2\) for \(n=1, 2, 3, \dots\)
This relation signifies that each term, starting from the second one, is calculated by adding 2 to the preceding term.

• For example, starting with \(a_0 = -4\):
• \(a_1 = -4 + 2 = -2\)
• Then, \(a_2 = -2 + 2 = 0\)
• And, \(a_3 = 0 + 2 = 2\)

The recurrence relation serves as the ongoing instruction manual that details how to proceed from one term to the next seamlessly, making it possible to generate the sequence to any desired term.

A general formula is derived to express any term in the sequence explicitly in terms of its position, \(n\). This bypasses the need to calculate every previous term, providing a direct way to find any specified term.
For this exercise, the relationship was identified through observation of the linear increase by 2 for each term. The formula can be expressed as:

• \(a_n = a_0 + 2n\) where \(n \geq 1\)
• Since \(a_0 = -4\), substituting it gives: \(a_n = -4 + 2n\)

This general formula is special because it allows you to compute any term in the sequence without reliance on all the preceding terms. For instance, if you're asked to find \(a_5\), simply plug 5 into the formula: \(a_5 = -4 + 2 \, \times \, 5 = 6\). Thus, the general formula is a powerful tool for simplifying and solving problems involving recursively defined sequences by providing a direct path to any term.