Problem 51 - College Algebra with Corequisite Support

Question

For the following exercises, graph the first five terms of the indicated sequence \(a_{n}=\frac{(n+1) !}{(n-1) !}\)

Step-by-Step Solution

Answer

The terms are 2, 6, 12, 20, and 30.

Step 1: Understanding the Problem

To solve this problem, we need to find and graph the first five terms of the sequence given by the formula \(a_{n}=\frac{(n+1)!}{(n-1)!}\). We will calculate these terms for \(n = 1\) to \(n = 5\).

Step 2: Calculate the First Term

Substitute \(n = 1\) into the formula: \(a_1 = \frac{(1+1)!}{(1-1)!} = \frac{2!}{0!}\). Recall that \(2! = 2 \times 1 = 2\) and by definition \(0! = 1\). Thus, \(a_1 = \frac{2}{1} = 2\).

Step 3: Calculate the Second Term

Substitute \(n = 2\) into the formula: \(a_2 = \frac{(2+1)!}{(2-1)!} = \frac{3!}{1!}\). Since \(3! = 3 \times 2 \times 1 = 6\) and \(1! = 1\), \(a_2 = \frac{6}{1} = 6\).

Step 4: Calculate the Third Term

Substitute \(n = 3\) into the formula: \(a_3 = \frac{(3+1)!}{(3-1)!} = \frac{4!}{2!}\). We know \(4! = 4 \times 3 \times 2 \times 1 = 24\) and \(2! = 2 \times 1 = 2\). Thus, \(a_3 = \frac{24}{2} = 12\).

Step 5: Calculate the Fourth Term

Substitute \(n = 4\) into the formula: \(a_4 = \frac{(4+1)!}{(4-1)!} = \frac{5!}{3!}\). With \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\) and \(3! = 3 \times 2 \times 1 = 6\), we have \(a_4 = \frac{120}{6} = 20\).

Step 6: Calculate the Fifth Term

Substitute \(n = 5\) into the formula: \(a_5 = \frac{(5+1)!}{(5-1)!} = \frac{6!}{4!}\). Here, \(6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720\) and \(4! = 4 \times 3 \times 2 \times 1 = 24\). Thus, \(a_5 = \frac{720}{24} = 30\).

Step 7: Graphing the Terms

The first five terms of the sequence are \(a_1 = 2\), \(a_2 = 6\), \(a_3 = 12\), \(a_4 = 20\), \(a_5 = 30\). To graph these terms, plot the points \((1, 2)\), \((2, 6)\), \((3, 12)\), \((4, 20)\), and \((5, 30)\) on a coordinate plane, with the term index \(n\) on the x-axis and the value \(a_n\) on the y-axis.

Key Concepts

Factorials

Factorials are mathematical functions denoted by the symbol "!" and used to find the product of all positive integers up to a certain number. For instance, the factorial of a number "n" is represented as "n!" and is calculated as the product of all positive integers from 1 to "n".

• For example, 4! (read as "four factorial") equals \[ 4 \times 3 \times 2 \times 1 = 24 \]
• An important aspect of factorials is that 0! is defined to be 1. This might seem counterintuitive at first, but it is a convention that helps in the simplification of mathematical calculations, especially in combinatorics and sequences.

Factorials grow very rapidly as numbers increase, and they are frequently used in algebra to express sequences compactly and to perform operations that involve combinations and permutations.

Graphing sequences

Graphing sequences is a great way to visually interpret the behavior of sequences over a range of terms. In the context of the given exercise, graphing is used to plot the calculated terms of the sequence on a coordinate plane.
Here's how you can graph a sequence effectively:

• Calculate the terms of the sequence for the values of interest. For instance, the first five terms of a sequence.
• Each term can be represented by a point define the x-coordinate as the index of the sequence (e.g., 1, 2, 3, etc.) and the y-coordinate as the value of the term.
• Plot these points on a chart where the horizontal axis (x-axis) represents the term index and the vertical axis (y-axis) represents the term value.

This graphical approach not only helps in understanding the growth pattern of the sequence but also gives a better insight into its trend.

Integer sequences

Integer sequences are sequences consisting of ordered lists of integers. They can be generated by certain rules or formulas. The sequence in our exercise is defined by the formula:\[ a_n = \frac{(n+1)!}{(n-1)!} \]This sequence generates terms based on factorial expressions, and each term has an index starting from 1 (or any other chosen starting point).
Integer sequences can have different characteristics:

• They may be arithmetic, geometric, or neither, depending on the rule defining them.
• Some sequences, such as the Fibonacci sequence, have special properties and applications in mathematics and other fields.
• Integer sequences can be finite or infinite, depending on whether they've been defined over a limited set of values or continue infinitely as the index increases.

Understanding integer sequences involves recognizing these patterns and how the formula generates each successive number in the sequence.

Term calculations

Term calculations involve finding specific values in a sequence by substituting index values into the rule or formula that defines the sequence. In the exercise problem, for each term, we substitute a value of "n" into the formula:\[ a_n = \frac{(n+1)!}{(n-1)!} \]
Let's calculate a few terms to fully grasp the process:

• For \( n = 1 \): Substitute "n" into the formula to find \( a_1 \).
• The calculation becomes \( a_1 = \frac{(1+1)!}{(1-1)!} = \frac{2!}{0!} \) which equals 2 because 0! = 1 by definition.
• Similarly, for \( n = 2 \), calculate \( a_2 = \frac{3!}{1!} \) resulting in 6.
• Continue this process for higher values of "n" to find the subsequent terms.

Each calculation delivers a specific term of the sequence and relies heavily on understanding both the rule defining the sequence and the mathematical concept of factorials used within it. This sequence grows at an increasing rate with incremented indices due to the nature of factorial calculations.