Problem 28 - College Algebra with Corequisite Support

Question

For the following exercises, write the first five terms of the sequence. \(a_{1}=-4, \quad a_{n}=\frac{a_{n-1}+2 n}{a_{n-1}-1}\)

Step-by-Step Solution

Answer

The first five terms are: -4, 0, -6, -2/7, and -68/9.

Step 1: Find the first term

We can directly take the first term from the given information. The sequence starts with \(a_1 = -4\).

Step 2: Calculate the second term

To find \(a_2\), use the formula \(a_n = \frac{a_{n-1} + 2n}{a_{n-1} - 1}\) with \(n = 2\):\[a_2 = \frac{a_1 + 2 \times 2}{a_1 - 1} = \frac{-4 + 4}{-4 - 1} = \frac{0}{-5} = 0\]So, \(a_2 = 0\).

Step 3: Calculate the third term

Use the formula to find \(a_3\) with \(n = 3\):\[a_3 = \frac{a_2 + 2 \times 3}{a_2 - 1} = \frac{0 + 6}{0 - 1} = \frac{6}{-1} = -6\]Therefore, \(a_3 = -6\).

Step 4: Calculate the fourth term

Find \(a_4\) by substituting when \(n = 4\):\[a_4 = \frac{a_3 + 2 \times 4}{a_3 - 1} = \frac{-6 + 8}{-6 - 1} = \frac{2}{-7} = -\frac{2}{7}\]Thus, \(a_4 = -\frac{2}{7}\).

Step 5: Calculate the fifth term

Now, find \(a_5\) using \(n = 5\):\[a_5 = \frac{a_4 + 2 \times 5}{a_4 - 1} = \frac{-\frac{2}{7} + 10}{-\frac{2}{7} - 1}\]First calculate the numerator: \[-\frac{2}{7} + 10 = \frac{-2}{7} + \frac{70}{7} = \frac{68}{7}\]Now for the denominator:\[-\frac{2}{7} - 1 = -\frac{2}{7} - \frac{7}{7} = -\frac{9}{7}\]Thus:\[a_5 = \frac{\frac{68}{7}}{-\frac{9}{7}} = \frac{68}{7} \times \frac{7}{-9} = \frac{68}{-9} = -\frac{68}{9}\]Therefore, \(a_5 = -\frac{68}{9}\).

Key Concepts

Arithmetic Sequences

Arithmetic sequences are a fundamental concept in mathematics, especially within the study of sequences and series. In an arithmetic sequence, each term after the first is generated by adding a constant, known as the common difference, to the previous term. This type of sequence is very predictable, as the difference between any two consecutive terms is always the same.

For instance, consider the arithmetic sequence: 2, 4, 6, 8, 10. Here, the common difference is 2 because each term is generated by adding 2 to the previous term. This predictable pattern allows us to easily calculate any term in the sequence if we know the first term and the common difference.

The formula for the nth term of an arithmetic sequence is given by:

• \( a_n = a_1 + (n - 1) imes d \)

where \(a_n\) is the nth term, \(a_1\) is the first term, and \(d\) is the common difference.

Recursive Sequences

Recursive sequences are an interesting type of mathematical sequences. They involve rules that tell you how to calculate a term using the preceding terms in the sequence. In other words, a recursive formula provides a way to generate the terms of the sequence sequentially, one after the other.

A common example of a recursive sequence is found in the famous Fibonacci sequence, where each term is the sum of the two preceding terms. However, recursive sequences can be much more complex and can involve operations beyond mere addition.

The original exercise involves a recursive sequence, given by the expression:

• \( a_n = \frac{a_{n-1} + 2n}{a_{n-1} - 1} \)

In this formula, \( a_{n-1} \) is the previous term, and \( n \) is the position of the term in the sequence. To find any term, you must first know the terms before it, which makes solving recursive sequences often challenging but also rewarding.

Terms of a Sequence

The terms of a sequence are simply the individual values that make up the sequence. Each term has a position, or an index, that indicates its place within the sequence. Understanding the terms in a sequence is crucial because they help us to identify patterns, which in turn can allow us to predict or calculate future terms.

In the given exercise, we started with the first term \(a_1 = -4\). Each subsequent term was calculated using the recursive formula provided. When working with such sequences, it's important to keep track of each term's position and the calculation method.

Calculating terms in sequences often involves repetition and careful evaluation of the formulas. As seen in the step-by-step solution, knowing the first term allowed us to determine the second, and so forth. This sequential calculation is vital in forming a clear understanding of how the sequence behaves and evolves over time.