Problem 60 - College Algebra with Corequisite Support

Question

Find the first ten terms of the sequence \(a_{1}=8\), \(a_{n}=\frac{\left(a_{n-1}+1\right) !}{a_{n-1} !}\).

Step-by-Step Solution

Answer

The first ten terms are 8, 9, 10, 11, 12, 13, 14, 15, 16, 17.

Step 1: Understanding the Formula

The sequence is given by the recurrence relation: \(a_1 = 8\), and \(a_n = \frac{(a_{n-1} + 1)!}{a_{n-1}!}\). This formula can be simplified to \(a_n = a_{n-1} + 1\) because the factorial in the numerator cancels all but one term in the denominator.

Step 2: Calculating the Second Term

Use the simplified formula: \(a_2 = a_1 + 1 = 8 + 1 = 9\).

Step 3: Calculating the Third Term

Continue using the formula: \(a_3 = a_2 + 1 = 9 + 1 = 10\).

Step 4: Calculating the Fourth Term

Keep applying the formula: \(a_4 = a_3 + 1 = 10 + 1 = 11\).

Step 5: Calculating the Fifth Term

Apply the formula again: \(a_5 = a_4 + 1 = 11 + 1 = 12\).

Step 6: Calculating the Sixth Term

Using the pattern: \(a_6 = a_5 + 1 = 12 + 1 = 13\).

Step 7: Calculating the Seventh Term

Continuing the sequence: \(a_7 = a_6 + 1 = 13 + 1 = 14\).

Step 8: Calculating the Eighth Term

Further application: \(a_8 = a_7 + 1 = 14 + 1 = 15\).

Step 9: Calculating the Ninth Term

Next term in the sequence: \(a_9 = a_8 + 1 = 15 + 1 = 16\).

Step 10: Calculating the Tenth Term

Final term needed: \(a_{10} = a_9 + 1 = 16 + 1 = 17\).

Key Concepts

Recurrence Relation

A recurrence relation is a formula that defines each term of a sequence using the preceding terms. In this exercise, the sequence begins with a first term, known as the initial condition, and then uses a rule to determine subsequent terms based on the previous ones. Specifically, the recurrence relation provided was: \( a_n = \frac{(a_{n-1} + 1)!}{a_{n-1}!} \). However, it simplifies to \( a_n = a_{n-1} + 1 \) upon closer examination, because the factorials in the numerator and denominator cancel out.

• The initial term is \( a_1 = 8 \).
• The relation shows how each term is developed from its predecessor by adding 1.

This makes it easy to predict or compute future terms since it follows a straightforward pattern. Recurrence relations can vary widely in their complexity, from very simple arithmetic steps like this one, to more intricate calculations requiring multiple terms and operations.

Factorials

Factorials are an important part of calculating sequences in certain algebra problems. A factorial, denoted as \( n! \), is the product of all positive integers from 1 up to \( n \). In mathematical terms:\[ n! = n \times (n-1) \times \cdots \times 2 \times 1\]For example, \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \).

• Factorials grow very rapidly. This can make calculations quite large very quickly.
• They are often used in permutations, combinations, and various other mathematical concepts.

In the exercise, the factorial expression \((a_{n-1} + 1)!\) in the numerator and \(a_{n-1}!\) in the denominator simplify to a small step due to the cancellation, reflecting how factorials can sometimes simplify complex expressions.

Arithmetic Sequence

An arithmetic sequence is a series of numbers in which each term after the first is generated by adding a constant difference to the preceding term. This sequence was identified in the exercise by the pattern \(a_n = a_{n-1} + 1\). Here, the constant difference was 1:

• The first term was set at 8 \((a_1 = 8)\).
• Every subsequent term increased by 1, creating a straightforward, linear pattern.

Such sequences are predictable as each term builds directly from the last, making them easy to analyze and compute by hand. Arithmetic sequences are commonly used in various mathematical contexts due to their simplicity and predictability, and solutions often involve identifying the first term and common difference.