Problem 66 - College Algebra with Corequisite Support

Question

List the first six terms of the sequence \(a_{n}=\frac{n !}{n}\).

Step-by-Step Solution

Answer

1, 1, 2, 6, 24, 120

Step 1: Understand the Sequence

The given sequence is defined by the formula \( a_n = \frac{n!}{n} \). \( n! \) stands for the factorial of \( n \), which is the product of all positive integers up to \( n \). The sequence is dependent on the integer \( n \).

Step 2: Calculate the First Term

For the first term, substitute \( n=1 \) into the formula: \( a_1 = \frac{1!}{1} = \frac{1}{1} = 1 \).

Step 3: Calculate the Second Term

For the second term, substitute \( n=2 \) into the formula: \( a_2 = \frac{2!}{2} = \frac{2 \cdot 1}{2} = 1 \).

Step 4: Calculate the Third Term

For the third term, substitute \( n=3 \) into the formula: \( a_3 = \frac{3!}{3} = \frac{3 \cdot 2 \cdot 1}{3} = 2 \).

Step 5: Calculate the Fourth Term

For the fourth term, substitute \( n=4 \) into the formula: \( a_4 = \frac{4!}{4} = \frac{4 \cdot 3 \cdot 2 \cdot 1}{4} = 6 \).

Step 6: Calculate the Fifth Term

For the fifth term, substitute \( n=5 \) into the formula: \( a_5 = \frac{5!}{5} = \frac{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{5} = 24 \).

Step 7: Calculate the Sixth Term

For the sixth term, substitute \( n=6 \) into the formula: \( a_6 = \frac{6!}{6} = \frac{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{6} = 120 \).

Step 8: List the First Six Terms

The first six terms of the sequence are: 1, 1, 2, 6, 24, and 120.

Key Concepts

Sequence Terms

A sequence in mathematics is an ordered list of numbers that follow a specific rule or pattern. When dealing with sequence terms, it is crucial to understand how each term is defined within the sequence.
For instance, in this exercise, we are working with the sequence defined by the formula \( a_n = \frac{n!}{n} \). Here, each term \( a_n \) is calculated using the integer \( n \), which changes as you move from one term to the next. This sequence specifically uses factorials in its formula, creating a pattern where each term builds upon the last.
By calculating the sequence terms, you can observe a growing progression of values, revealing how the sequence develops. For each \( n \), you perform the operation of taking the factorial of \( n \) and then dividing by \( n \) itself. This pattern is what distinguishes the sequence and allows one to predict subsequent terms from its calculated progression.

Factorials

Factorials are a fundamental concept in mathematics, represented by the symbol \(!\). The factorial of a positive integer \( n \), denoted \( n! \), is the product of all positive integers from 1 up to \( n \).

• \( 1! = 1 \)
• \( 2! = 2 \times 1 = 2 \)
• \( 3! = 3 \times 2 \times 1 = 6 \)
• \( 4! = 4 \times 3 \times 2 \times 1 = 24 \)
• \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \)
• And so on...

Factorials grow very quickly, and they play a critical role in many areas of mathematics, including permutations, combinations, and of course, sequences. In our specific exercise, factorials help define the pattern of the sequence terms \( a_n = \frac{n!}{n} \), demonstrating how each term within our sequence is constructed.

Series Calculations

Series calculations often involve computing terms of a sequence based on a defining formula, and then possibly summing them up for deeper analysis. In the context of our exercise, we focus on computing each term in series, following the formula formula \( a_n = \frac{n!}{n} \).

• The first term is calculated by evaluating \( \frac{1!}{1} \), which equals 1.
• The second term uses \( \frac{2!}{2} \), also resulting in a value of 1.
• The third term is \( \frac{3!}{3} \), producing 2.
• The fourth term derives from \( \frac{4!}{4} \), equaling 6.
• The fifth term is \( \frac{5!}{5} \), coming out to 24.
• Finally, the sixth term employs \( \frac{6!}{6} \), resulting in 120.

These calculations create our sequence of terms: 1, 1, 2, 6, 24, and 120. Understanding each step in calculating these terms helps reinforce the principles of factorials and illustrates how sequences and series form a bridge between simple arithmetic operations and more complex mathematical analysis.