Problem 17 - College Algebra with Corequisite Support

Question

For the following exercises, write the first eight terms of the piecewise sequence. \(a_{n}=\left\\{\begin{array}{ll}\frac{n^{2}}{2 n+1} & \text { if } n \quad \leq 5 \\ n^{2}-5 & \text { if } n>5\end{array}\right.\)

Step-by-Step Solution

Answer

The first eight terms are \(\frac{1}{3}, \frac{4}{5}, \frac{9}{7}, \frac{16}{9}, \frac{25}{11}, 31, 44, 59\).

Step 1: Understanding the Sequence

This sequence is defined piecewise, which means it has different formulas for different values of \(n\). We need to write the first eight terms of the sequence using these formulas.

Step 2: Apply the First Piecewise Formula for \(n \leq 5\)

For terms where \(n \leq 5\), we use the formula \(a_n = \frac{n^2}{2n+1}\). We will calculate the sequence terms for \(n = 1, 2, 3, 4,\) and \(5\).1. \(a_1 = \frac{1^2}{2 \times 1 + 1} = \frac{1}{3}\)2. \(a_2 = \frac{2^2}{2 \times 2 + 1} = \frac{4}{5}\)3. \(a_3 = \frac{3^2}{2 \times 3 + 1} = \frac{9}{7}\)4. \(a_4 = \frac{4^2}{2 \times 4 + 1} = \frac{16}{9}\)5. \(a_5 = \frac{5^2}{2 \times 5 + 1} = \frac{25}{11}\)

Step 3: Apply the Second Piecewise Formula for \(n > 5\)

For terms where \(n > 5\), we use the formula \(a_n = n^2 - 5\). We will calculate the sequence terms for \(n = 6, 7,\) and \(8\).6. \(a_6 = 6^2 - 5 = 36 - 5 = 31\)7. \(a_7 = 7^2 - 5 = 49 - 5 = 44\)8. \(a_8 = 8^2 - 5 = 64 - 5 = 59\)

Step 4: Write Down the First Eight Terms

From the calculations, the first eight terms of the sequence are:1. \(a_1 = \frac{1}{3}\)2. \(a_2 = \frac{4}{5}\)3. \(a_3 = \frac{9}{7}\)4. \(a_4 = \frac{16}{9}\)5. \(a_5 = \frac{25}{11}\)6. \(a_6 = 31\)7. \(a_7 = 44\)8. \(a_8 = 59\)

Key Concepts

Sequence Terms

To understand a sequence, it's essential to know what sequence terms are. A sequence is an ordered list of numbers following a specific pattern, called terms. Each number in a sequence is a term, and they are usually denoted as \(a_1, a_2, a_3, \ldots \) for the first, second, third terms, and so on. In the context of our exercise, we are asked to find the first eight terms of a piecewise sequence.
- The first term in our sequence is \(a_1 = \frac{1}{3}\).
- The second term is \(a_2 = \frac{4}{5}\), and so on.
These terms are calculated based on the formula given for each part of the piecewise sequence, which changes depending on the value of \(n\). Understanding how to compute and arrange these terms is crucial in both identifying patterns and solving sequence-related questions.

Piecewise Function

A piecewise function is a function that has different expressions based on different conditions. In mathematical terms, it's a rule that assigns a number to each input, which might depend on which interval or condition that input belongs to.

Understanding Piecewise Functions

In our exercise, the sequence is defined piecewise. This means we use different formulas depending on whether \(n\) is less than or equal to 5, or greater than 5.
- For \(n \leq 5\), we use the formula \(a_n = \frac{n^2}{2n+1}\).
- For \(n > 5\), we use the formula \(a_n = n^2 - 5\).
Different scenarios require us to plug into different parts of the function, allowing flexibility and control over how outputs are calculated based on given inputs. Understanding this concept helps in effectively handling and breaking down complex sequences into manageable formulas, ensuring each portion is handled correctly.

Algebraic Sequences

Algebraic sequences are sequences where terms are determined using algebraic expressions. These expressions often involve variables, coefficients, and constants. In the case of our piecewise sequence, algebra helps us find and simplify the terms according to the rules set out by the piecewise function for varying values of \(n\).

Calculating Terms in Algebraic Sequences

For our sequence, consider how the expressions \(\frac{n^2}{2n+1}\) and \(n^2 - 5\) are used. Each represents an algebraic expression determining the terms of the sequence:
- For \(1 \leq n \leq 5\), the expression is \(\frac{n^2}{2n+1}\), which involves dividing a quadratic expression by a linear expression.
- For \(n > 5\), the expression is \(n^2 - 5\), a simple quadratic expression.
This approach shows that algebra not only helps in creating sequences but in understanding their behavior and form, allowing us to predict and evaluate terms efficiently.