Problem 27 - College Algebra with Corequisite Support

Question

For the following exercises, write the first five terms of the sequence. \(a_{1}=3, \quad a_{n}=(-3) a_{n-1}\)

Step-by-Step Solution

Answer

The terms are 3, -9, 27, -81, and 243.

Step 1: Understand the Problem

We need to find the first five terms of the sequence where the first term is given as \( a_1 = 3 \) and each subsequent term is calculated using the formula \( a_n = (-3) a_{n-1} \).

Step 2: Find the Second Term

To find the second term \( a_2 \), we use the given formula: \( a_2 = (-3) a_1 \). Since \( a_1 = 3 \), we have \( a_2 = (-3) \times 3 = -9 \).

Step 3: Find the Third Term

Using the formula \( a_3 = (-3) a_2 \), and substituting \( a_2 = -9 \), we get \( a_3 = (-3) \times (-9) = 27 \).

Step 4: Find the Fourth Term

For the fourth term, apply the formula: \( a_4 = (-3) a_3 \). Substituting \( a_3 = 27 \), we calculate \( a_4 = (-3) \times 27 = -81 \).

Step 5: Find the Fifth Term

Finally, calculate the fifth term using \( a_5 = (-3) a_4 \). With \( a_4 = -81 \), we have \( a_5 = (-3) \times (-81) = 243 \).

Key Concepts

Recursive Formulas

A recursive formula is a valuable tool in mathematics, particularly in sequences and series. It's a way to define a sequence by expressing each term as a function of the preceding term, rather than using a fixed equation. This approach is beneficial when there's a need to determine terms gradually, based on previous values.

In the given sequence problem, the recursive formula is defined as \( a_n = (-3) a_{n-1} \). Here, each term \( a_n \) relies on its predecessor \( a_{n-1} \). By starting with a known initial term, \( a_1 = 3 \), the formula allows us to compute all subsequent terms by continuing to apply the same operation repeatedly.

Exploring recursive formulas enhances understanding of relationships within sequences. It emphasizes the pattern or rule dictating how each term connects to the next, providing a clear strategy to work through sequences step-by-step. Recursive sequences are commonly encountered in various mathematical and real-world applications, making them essential to grasp.

Sequence Terms

Understanding sequence terms involves identifying each element generated in sequence as defined by a rule or formula. Sequence terms can evolve through various relationships: arithmetic progression, geometric progression, and others.

In this exercise, the sequence is produced by multiplying each term with a constant multiplier of \(-3\). It forms a geometric sequence, characterized by exponential growth or decay, depending on the multiplier's value. The first five terms of the sequence are quickly calculated step-by-step:

• First term \(a_1 = 3\)
• Second term \(a_2 = -9\)
• Third term \(a_3 = 27\)
• Fourth term \(a_4 = -81\)
• Fifth term \(a_5 = 243\)

The use of recursive formulas makes identifying sequence terms straightforward, assuming an initial term (or terms) is provided.

Grasping the structure of geometric sequences prepares students for more advanced mathematical concepts and empowers them to predict sequence behavior over larger scopes efficiently.

Problem Solving in Algebra

Problem solving in algebra often requires combining various ideas and methodologies to arrive at a solution. In this context, recognizing patterns through sequences and utilizing recursive formulas are tools which make algebraic problem solving not only manageable but profoundly satisfying.

In our exercise, the stepwise approach mirrors an algebraic reasoning process:

• Identify what is given: Initial term and recursive relationship.
• Apply the recursive formula iteratively to discover unknown terms.
• Check the calculations at each step for accuracy, ensuring the logic holds through the sequence.

This blend of procedural and conceptual understanding allows for efficient problem resolution.
Furthermore, such problems illustrate the importance of logical reasoning in algebra. They encourage students to engage systematically, ensuring each step logically leads to the next. Mastery in problem-solving within algebra boosts confidence in tackling complex scenarios, both mathematical and real-world.