Problem 53 - Precalculus with Limits

Question

In Exercises 47-62, write an expression for the apparent \(n\)th term of the sequence. (Assume that \( n \) begins with 1.) \( \dfrac{2}{1}, \dfrac{3}{3}, \dfrac{4}{5}, \dfrac{5}{7}, \dfrac{6}{9}, \dots \)

Step-by-Step Solution

Answer

The nth term of the sequence is \(\dfrac{n+1}{2n-1}\).

Step 1: Identify the Numerator

Observe the numerators of the sequence. They start from 2 (for n=1) and increase by 1 for each term i.e. for nth term numerator is \(n+1\).

Step 2: Identify the Denominator

Now observe the denominators of the sequence. They start from 1 (for n=1) and increase by 2 for each term i.e. for nth term denominator is \(2n-1\).

Step 3: Combine the Numerator and Denominator

Combine the numerator and denominator to form the nth term. So, the expression for the apparent nth term of the sequence is: \(\dfrac{n+1}{2n-1}\).

Key Concepts

Sequence Pattern

Sequence patterns are fundamental in understanding and predicting the next elements in a numerical or algebraic series. When analyzing sequence patterns, we discern the regular intervals or operations that define the sequence. For the exercise at hand, the sequence begins with fractions: \(\frac{2}{1}, \frac{3}{3}, \frac{4}{5}, \frac{5}{7}, \frac{6}{9}, \dots\) which clearly show a regular pattern.

To decode this, we see that the numerators are consecutive numbers beginning at 2 like 2, 3, 4, 5, 6, following an arithmetic sequence, where each term increases by 1. If we relate this to the sequence position \( n \), the pattern for the numerator can be described by the expression \( n+1 \).

Similarly, the denominators are progressing as 1, 3, 5, 7, 9, forming an arithmetic sequence as well, with the first term being 1 and each subsequent term increasing by 2. This pattern is matched by the expression \( 2n-1 \). Thus, understanding sequence patterns is about decomposition and identification of such relations.

Numerator and Denominator

In fractions, the numerator and the denominator serve different roles that are crucial in both the structure and value of a fraction.

The **numerator** is the top part of the fraction. In our sequence, it essentially counts or denotes the term's place in the sequence, modified by an arithmetic rule. From the given progression of numerators: 2, 3, 4, 5, 6, we identify a clear pattern, where each term can be described as \( n+1 \). This is because each numerator is one more than its sequential position \( n \).

The **denominator** in this series reveals a different pattern, beginning with 1 and increasing by 2 for each succeeding term. Observed as 1, 3, 5, 7, 9, it can be expressed as \( 2n-1 \). Each denominator is twice its position in the sequence minus 1, illustrating how frequent a simple increment can interplay with sequence logic.

By identifying both numerical patterns, we can understand and predict further fractions in the sequence.

Mathematical Expression Writing

Writing mathematical expressions effectively means converting patterns or relationships into algebraic formulations. This is particularly useful in describing sequences, as it offers a streamlined representation.

In this exercise, we're tasked with creating an expression for the \( n \)th term of our fractional sequence. Based on the identified patterns:

• Numerator given by \( n+1 \)
• Denominator given by \( 2n-1 \)

we can seamlessly combine these to write the expression for any term \( a_n \): \( \frac{n+1}{2n-1} \).

Such expressions allow anyone to substitute any integer \( n \) to find the specific term in the sequence directly, highlighting the power of expressions to simplify and encapsulate complex relations.

This process of transforming observed numerical patterns into generalized mathematical statements is foundational in both constructing and understanding the language of mathematics.