Problem 27 - Beginning and Intermediate Algebra

Question

Find a formula for the general term, \(a_{n},\) of each sequence. $$\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \ldots$$

Step-by-Step Solution

Answer

The formula for the general term of the sequence is: \(a_n = \frac{n}{n+1}\).

Step 1: Determine the pattern in the sequence by analyzing the terms

For each term in the sequence, we can see that the numerator is one less than the denominator, which are both consecutive integers. So, we can represent the general term \(a_n\) as a fraction whose numerator is \(n\) and the denominator is \(n + 1\), i.e., $$a_n = \frac{n}{n+1}$$

Step 2: Test the formula with the first few terms of the sequence.

To verify if the formula derived above is correct, let's test it with the first few terms of the sequence: Term 1: \(a_1 = \frac{1}{1 + 1} = \frac{1}{2}\) Term 2: \(a_2 = \frac{2}{2 + 1} = \frac{2}{3}\) Term 3: \(a_3 = \frac{3}{3 + 1} = \frac{3}{4}\) As we can see, the formula we derived matches the terms in the given sequence.

Step 3: Write the formula for the general term of the sequence

Now that we have confirmed the formula works for the first few terms, we can write the general formula for the \(n^{th}\) term of the sequence as: $$a_n = \frac{n}{n+1}$$

Key Concepts

General Term Formula

When working with sequences, recognizing and establishing a general term formula is crucial. It allows us to express any term within the sequence as a function of its position (or index), denoted usually by \( n \). In our exercise, we're given the sequence \( \frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \ldots \). What stands out is the simple relationship between each term's numerator and denominator. For each term \( a_n \), the numerator is equal to its index \( n \) and the denominator is one more than the numerator, \( n + 1 \). Therefore, the general formula representing each term in this sequence is:

• \( a_n = \frac{n}{n+1} \)

This formula gives anyone analyzing the sequence the power to find any \( n^{th} \) term quickly by simply plugging the number corresponding to the position in place of \( n \). It's a handy shortcut instead of manually extending the sequence and confirms the pattern effortlessly.

Sequence Pattern Recognition

Recognizing patterns within sequences is a skill that lies at the heart of understanding them. It involves identifying repeated mathematical relationships or regularities between sequence terms. For this sequence, let's identify the distinctive pattern:

• Numerator is the term's position, i.e., the consecutive natural numbers \( 1, 2, 3, \ldots \).
• Denominator is the numerator increased by one, forming \( 2, 3, 4, \ldots \).

This pattern is both simple and elegant. It's a linear relationship between the numerator and the denominator, which is typical for many known sequences. Recognizing such patterns helps us derive the general formula efficiently, as seen in this case where \( a_n = \frac{n}{n+1} \). Understanding pattern recognition in sequences allows us to make predictions and form logical extensions for any given sequence.

Testing Formula Validity

Once a general term formula is derived, testing its validity is important. This ensures that the formula applies correctly to the sequence's terms. This step involves calculating the first few terms using the derived formula and comparing them against the original sequence terms. Let's revisit the initial terms using our formula \( a_n = \frac{n}{n+1} \):

• For \( n = 1 \), \( a_1 = \frac{1}{1+1} = \frac{1}{2} \) matches the first term.
• For \( n = 2 \), \( a_2 = \frac{2}{2+1} = \frac{2}{3} \) matches the second term.
• For \( n = 3 \), \( a_3 = \frac{3}{3+1} = \frac{3}{4} \) matches the third term.

Each computation confirms the accuracy of our general term formula, demonstrating its reliability for the sequence. Testing is an essential part of sequence analysis, as it validates that all terms of the sequence are accurately represented by the formula.