Problem 31 - College Algebra

Question

Write the first eight terms of the sequence. $$a_{1}=\frac{1}{24}, a_{2}=1, a_{n}=\left(2 a_{n-2}\right)\left(3 a_{n-1}\right)$$

Step-by-Step Solution

Answer

The first eight terms are \( \frac{1}{24}, 1, \frac{1}{4}, \frac{3}{2}, \frac{9}{4}, \frac{81}{4}, \frac{2187}{8}, \frac{59049}{8} \).

Step 1: Write Down Given Terms

We start by noting the given terms of the sequence. The first term is \( a_1 = \frac{1}{24} \) and the second term is \( a_2 = 1 \).

Step 2: Apply Recursive Formula for Third Term

For the third term, use the recursive formula: \( a_3 = (2a_{1})(3a_{2}) \). Substituting the known values, \( a_3 = (2 \times \frac{1}{24})(3 \times 1) = \frac{1}{12} \times 3 = \frac{1}{4} \).

Step 3: Calculate Fourth Term

For the fourth term, use \( a_4 = (2a_{2})(3a_{3}) \). Substitute \( a_2 \) and \( a_3 \): \( a_4 = (2 \times 1)(3 \times \frac{1}{4}) = 2 \times \frac{3}{4} = \frac{3}{2} \).

Step 4: Calculate Fifth Term

For the fifth term, apply \( a_5 = (2a_{3})(3a_{4}) \). With \( a_3 = \frac{1}{4} \) and \( a_4 = \frac{3}{2} \), \( a_5 = (2 \times \frac{1}{4})(3 \times \frac{3}{2}) = \frac{1}{2} \times \frac{9}{2} = \frac{9}{4} \).

Step 5: Calculate Sixth Term

For the sixth term, use \( a_6 = (2a_{4})(3a_{5}) \). Plugging in \( a_4 = \frac{3}{2} \) and \( a_5 = \frac{9}{4} \), \( a_6 = (2 \times \frac{3}{2})(3 \times \frac{9}{4}) = 3 \times \frac{27}{4} = \frac{81}{4} \).

Step 6: Calculate Seventh Term

For \( a_7 \), use \( a_7 = (2a_{5})(3a_{6}) \). Substitute \( a_5 = \frac{9}{4} \) and \( a_6 = \frac{81}{4} \), thus \( a_7 = (2 \times \frac{9}{4})(3 \times \frac{81}{4}) = \frac{9}{2} \times \frac{243}{4} = \frac{2187}{8} \).

Step 7: Calculate Eighth Term

For \( a_8 \), use \( a_8 = (2a_{6})(3a_{7}) \). Substitute \( a_6 = \frac{81}{4} \) and \( a_7 = \frac{2187}{8} \), giving \( a_8 = (2 \times \frac{81}{4})(3 \times \frac{2187}{8}) = \frac{162}{4} \times \frac{6561}{8} = \frac{59049}{8} \).

Key Concepts

sequence terms

In mathematics, a sequence is an ordered list of numbers. Each number in this list is called a term. Sequence terms are typically denoted by lowercase letters with a subscript indicating their position in the sequence, such as \( a_1, a_2, a_3, \ldots \). This subscript helps to easily refer to or work with specific terms when analyzing a sequence.

When dealing with sequences, especially in exercises like the one given, it is crucial to understand each term's role. Sometimes sequences have only a few terms provided, while the rest need to be calculated using certain rules or formulas. For example, in this exercise, the first two terms of the sequence are given as \( a_1 = \frac{1}{24} \) and \( a_2 = 1 \), making them the starting point for further calculations.

Knowing the initial terms allows us to apply formulas or recursive rules to determine subsequent terms. Understanding the structure and basis of sequence terms is fundamental, as it ensures that any developed pattern or rule holds true throughout the entire sequence.

recurrence relation

A recurrence relation is a mathematical equation that relates terms in a sequence, making it possible to calculate the entire sequence from initial terms. Recurrence relations are particularly useful in computer science, mathematics, and various scientific fields, as they allow complex sequences to be described compactly and calculated progressively.

In the exercise provided, the sequence is governed by the recurrence relation \( a_n = (2a_{n-2})(3a_{n-1}) \). This formula expresses each term \( a_n \) in terms of the two preceding terms, namely \( a_{n-2} \) and \( a_{n-1} \). With this recursive relationship, each new term is calculated using the two terms before it, applying a consistent algebraic transformation.

To use a recurrence relation effectively:

• Identify the base case: Begin with the given starting terms, as seen earlier with \( a_1 \) and \( a_2 \).
• Substitute into the relation: Iteratively plug these terms into the recurrence formula to find the next term, one by one.
• Repeat the process: Continue substituting with newly calculated terms to populate the sequence further.

Recurrence relations are a vital concept that link sequence terms together systematically and are foundational to solving complex sequence problems.

algebraic sequences

Algebraic sequences are sequences whose terms are derived using algebraic expressions or formulas. These can involve various operations like addition, subtraction, multiplication, division, or more complex manipulations. The exercise at hand utilizes an algebraic sequence where the terms are generated through a prescribed formula repeatedly applied to earlier terms.

Unlike arithmetic or geometric sequences that follow simple patterns of addition or multiplication of a constant, algebraic sequences can encompass a broader array of operations. For instance, in this exercise, the formula \( a_n = (2a_{n-2})(3a_{n-1}) \) involves both multiplication by constants and use of prior terms, thus introducing more complexity than linear sequences.

Understanding algebraic sequences involves:

• Grasping the formula: Clearly understanding the sequence's defining algebraic expression or equation.
• Consistent application: Accurately applying the formula across all terms to ensure every term fits the sequence rule.
• Pattern identification: Recognizing emerging patterns that help predict and verify further terms.

Algebraic sequences challenge us to apply algebraic thinking dynamically, making them a fascinating area of study with practical implications across fields such as engineering, computing, and physics.