Problem 43 - College Algebra

Question

For the following exercises, write the first four terms of the sequence. $$ a_{n}=\frac{n !}{n^{2}} $$

Step-by-Step Solution

Answer

The first four terms are 1, \( \frac{1}{2} \), \( \frac{2}{3} \), \( \frac{3}{2} \).

Step 1: Evaluate the formula for n=1

Substitute 1 into the formula: \( a_{1} = \frac{1!}{1^2} = \frac{1}{1} = 1 \). Therefore, the first term is 1.

Step 2: Evaluate the formula for n=2

Substitute 2 into the formula: \( a_{2} = \frac{2!}{2^2} = \frac{2 \times 1}{4} = \frac{2}{4} = \frac{1}{2} \). Therefore, the second term is \( \frac{1}{2} \).

Step 3: Evaluate the formula for n=3

Substitute 3 into the formula: \( a_{3} = \frac{3!}{3^2} = \frac{3 \times 2 \times 1}{9} = \frac{6}{9} = \frac{2}{3} \). Therefore, the third term is \( \frac{2}{3} \).

Step 4: Evaluate the formula for n=4

Substitute 4 into the formula: \( a_{4} = \frac{4!}{4^2} = \frac{4 \times 3 \times 2 \times 1}{16} = \frac{24}{16} = \frac{3}{2} \). Therefore, the fourth term is \( \frac{3}{2} \).

Key Concepts

Factorial

The concept of a factorial is exciting and unique in mathematics. It is denoted by an exclamation mark (!). The factorial of a non-negative integer, say \( n \), is the product of all positive integers less than or equal to \( n \). So, for instance:

• \( 0! = 1 \) by definition.
• \( 1! = 1 \)
• \( 2! = 2 \times 1 = 2 \)
• \( 3! = 3 \times 2 \times 1 = 6 \)
• \( 4! = 4 \times 3 \times 2 \times 1 = 24 \)

Factorials grow very quickly with larger numbers. They are useful in many areas, such as permutations and combinations, describing sequences, and more. Understanding factorials helps in evaluating expressions like the one used in your exercise.

Algebraic Expressions

Algebraic expressions are mathematical phrases that can contain numbers, variables, and operation symbols. In your exercise, the algebraic expression is \( \frac{n!}{n^{2}} \). It combines factorials and exponents.

• The numerator, \( n! \), represents the factorial.
• The denominator, \( n^{2} \), represents the square of \( n \).

By substituting values into this expression, you can evaluate it to find specific terms of a sequence. Understanding how to manipulate algebraic expressions is crucial to solving math problems efficiently.

Step-by-Step Problem Solving

Solving math problems step-by-step is a powerful method to tackle seemingly complex problems. It allows you to break down a problem into manageable parts. In this exercise, each step represents evaluating the expression for different values of \( n \):

• Start with \( n = 1 \), evaluate the expression, and record the result.
• Repeat the process for \( n = 2, 3, \) and \( 4 \).

This systematic approach ensures that nothing is overlooked, and each computation builds on the previous one. It's a technique that can be applied across various mathematical disciplines and problem types, enhancing both understanding and precision in problem-solving.

Evaluating Sequences

Evaluating sequences involves finding specific terms within a sequence as defined by a formula. In your exercise, the formula \( a_{n} = \frac{n!}{n^{2}} \) was given to write out the first four terms.

• By substituting \( n = 1 \), the first term \( a_1 \) was found as 1.
• For \( n = 2 \), the second term \( a_2 \) is \( \frac{1}{2} \).
• With \( n = 3 \), the third term \( a_3 \) is \( \frac{2}{3} \).
• Finally, substituting \( n = 4 \) yields the fourth term \( a_4 \) as \( \frac{3}{2} \).

Each calculation provides a snapshot of the sequence, illustrating how each term relates to \( n \). Evaluating sequences is an important skill in mathematics, as it connects formulas to tangible values and helps students explore patterns and relationships within numbers.