Problem 27 - Algebra 2 and Trigonometry

Question

In \(19-30 :\) a. Write an algebraic expression that represents \(a_{n}\) for each sequence. b. Find the ninth term of each sequence. $$ \frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \dots $$

Step-by-Step Solution

Answer

The ninth term is \( \frac{9}{10} \).

Step 1: Analyze the Sequence

The sequence is \( \frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \ldots \). Notice that the numerator increases by 1 each term, starting from 1, and the denominator also increases by 1, starting from 2. This suggests the general term for the sequence is formed by \( a_n = \frac{n}{n+1} \).

Step 2: Write the General Formula

From the analysis of the sequence, we see that the pattern for the terms is \( \frac{n}{n+1} \). Therefore, the algebraic expression representing the \( n \)-th term is \( a_n = \frac{n}{n+1} \).

Step 3: Substitute to Find the Ninth Term

To find the ninth term \( a_9 \), substitute \( n = 9 \) into the formula \( a_n = \frac{n}{n+1} \). This gives \( a_9 = \frac{9}{9+1} = \frac{9}{10} \).

Key Concepts

Sequences and Series

A sequence is an ordered list of numbers following a specific pattern or rule. In this exercise, the sequence given is a list of fractions: \( \frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5} \ldots \). This is an example of an arithmetic sequence in the numerators and denominators, meaning each part increases by a fixed amount as you move down the line. Here, you can observe that both the numerator and the denominator increase by 1 with each subsequent term.

When you list out such numbers, you transform them into a mathematical structure called a series, if you decide to sum them up. However, in this exercise, we stick with a sequence. Identifying the pattern in a sequence is crucial, as it allows us to write a general formula to find any term in the sequence without listing the terms explicitly.

Numerator and Denominator

In fractions, the **numerator** is the top number, and the **denominator** is the bottom number. In the sequence provided, each numerator starts at 1 and increases sequentially by 1, while each denominator starts at 2 and follows the same pattern of increment.

Understanding these elements' roles is vital for creating the general formula. In our sequence, observe how these relate to their position:

• \(1^{st}\) term: Numerator = 1, Denominator = 2
• \(2^{nd}\) term: Numerator = 2, Denominator = 3
• \(n^{th}\) term: Numerator = n, Denominator = n+1

This systematic approach to interpreting numerators and denominators forms the backbone of our formula.

Formula Derivation

Deriving a formula from a sequence involves recognizing a consistent pattern in the sequence, then expressing it algebraically. In this exercise, we aim to discover how we can predict the \( n \)-th term of the sequence: \( \frac{n}{n+1} \).

The derivation process goes like this:

• Observe the increments in the sequence terms, both numerator and denominator rise by 1 in each step.
• Identify the position's influence: For the \( n \)-th term, numerators equal \( n \), and denominators are \( n+1 \).
• Check the consistency with several terms to assure the formula works universally within the sequence.

This understanding lets us forecast specific terms within the sequence without listing them, a powerful tool when working with sequences and series.