Problem 44 - College Algebra

Question

For the following exercises, write the first four terms of the sequence. $$ a_{n}=\frac{3 \cdot n !}{4 \cdot n !} $$

Step-by-Step Solution

Answer

The first four terms of the sequence are \( \frac{3}{4}, \frac{3}{4}, \frac{3}{4}, \frac{3}{4} \).

Step 1: Understand the Formula

The sequence is given by the formula \( a_n = \frac{3 \cdot n!}{4 \cdot n!} \). You are asked to find the first four terms of this sequence.

Step 2: Simplify the Formula

Simplify the expression \( \frac{3 \cdot n!}{4 \cdot n!} \). Since the \( n! \) in the numerator and the denominator cancel out, the expression simplifies to \( \frac{3}{4} \). Thus, \( a_n = \frac{3}{4} \) for any positive integer \( n \).

Step 3: Compute the First Term

To find the first term \( a_1 \), plug in \( n = 1 \) into the simplified formula: \( a_1 = \frac{3}{4} \).

Step 4: Compute the Second Term

For the second term \( a_2 \), use the simplified formula with \( n = 2 \): \( a_2 = \frac{3}{4} \).

Step 5: Compute the Third Term

For the third term \( a_3 \), use the simplified formula with \( n = 3 \): \( a_3 = \frac{3}{4} \).

Step 6: Compute the Fourth Term

For the fourth term \( a_4 \), use the simplified formula with \( n = 4 \): \( a_4 = \frac{3}{4} \).

Key Concepts

Factorials

Factorials are an essential aspect of sequences and series, especially when dealing with mathematical formulas. The factorial of a non-negative integer \( n \), denoted as \( n! \), is the product of all positive integers less than or equal to \( n \). For example, \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \). Factorials are useful for counting and permutations in mathematics.
In the given exercise, we see \( n! \) in both the numerator and denominator of the sequence formula. This notation facilitates simplifying expressions, especially when factorials cancel each other out.

Simplification

Simplification is a crucial step in solving mathematical problems, as it reduces complex expressions to their simplest form. In our exercise, the initial sequence formula is \( a_n = \frac{3 \cdot n!}{4 \cdot n!} \).
By recognizing that \( n! \) appears in both the numerator and denominator, you can cancel these terms out, essentially leaving us with an expression that no longer depends on \( n \). This results in a simplified, constant expression \( \frac{3}{4} \). Simplification makes calculation easier and less prone to error.

Term Calculation

Term calculation involves finding specific terms within a sequence. With the simplified expression \( a_n = \frac{3}{4} \), determining the terms of this sequence becomes straightforward. Regardless of the value of \( n \), \( a_n \) remains as \( \frac{3}{4} \).
To find the first four terms:

• First term \( (a_1) \): Set \( n = 1 \), resulting in \( a_1 = \frac{3}{4} \).
• Second term \( (a_2) \): Set \( n = 2 \), resulting in \( a_2 = \frac{3}{4} \).
• Third term \( (a_3) \): Set \( n = 3 \), resulting in \( a_3 = \frac{3}{4} \).
• Fourth term \( (a_4) \): Set \( n = 4 \), resulting in \( a_4 = \frac{3}{4} \).

Overall, the calculation shows that this sequence is constant, with all terms being the same value of \( \frac{3}{4} \). This method is efficient for sequences where simplification leads to a uniform result across different terms.