Problem 38 - Precalculus: Mathematics for Calculus

Question

Find the \(n\)th term of a sequence whose first several terms are given. \(1, \frac{1}{2}, 3, \frac{1}{4}, 5, \frac{1}{6}, \dots\)

Step-by-Step Solution

Answer

The nth term is given by \(a_n = 2k-1\) if odd, and \(a_n = \frac{1}{2k}\) if even.

Step 1: Analyze the Pattern of Integer Terms

Look for a consistent pattern among the integer terms in the sequence: 1, 3, 5, ... These appear to be increasing by 2 each time, which is an arithmetic sequence with a common difference of 2. The formula for the general term of this sequence is given by the expression for odd numbers: \(a_n = 2n - 1\).

Step 2: Analyze the Pattern of Fractional Terms

Now, analyze the pattern of the fractional terms: \(\frac{1}{2}, \frac{1}{4}, \frac{1}{6},...\). These denominators (2, 4, 6, ...) constitute an arithmetic sequence with a difference of 2, resulting in terms given by \(2n\). Thus, the fractional term follows the pattern \(\frac{1}{2n}\).

Step 3: Define Odd and Even Terms in the Sequence

The sequence alternates between an integer term and a fractional term. When \(n\) is odd, the term is integer and follows the formula \(2n - 1\). When \(n\) is even, the term is a fraction and follows \(\frac{1}{2n}\). This alternative pattern helps us create two separate functions depending on whether \(n\) is odd or even.

Step 4: Derive the General Formula for Any Term

Combine the observations from Steps 1 and 2 into a single expression that handles both odd and even \(n\). Using mathematical representation, the nth term \(a_n\) of the sequence can be defined as:\[ a_n = \begin{cases} 2k-1, & \text{if } n = 2k - 1 \ \frac{1}{2k}, & \text{if } n = 2k \end{cases} \] where \(k\) is the sequence index for integers (1, 2, 3, ...) corresponding to odd or even positions.

Key Concepts

Odd Numbers and Arithmetic Sequences

Odd numbers are numbers that cannot be evenly divided by 2. Examples include 1, 3, 5, 7, and so on. In mathematics, these numbers form a very specific pattern called an arithmetic sequence. An arithmetic sequence is a sequence of numbers with a constant difference between consecutive terms. In the sequence of odd numbers, the common difference is 2. This means each term is obtained by adding 2 to the previous term.

To find the general term of an arithmetic sequence, especially odd numbers, we use the formula:\[ a_n = 2n - 1 \]Here, \(a_n\) represents the nth term. The term \(2n - 1\) is derived from recognizing that each consecutive odd number is achieved by multiplying the sequence's position \(n\) by 2 and then subtracting 1. This formula is our guide to understanding where each odd number stands in its sequence.

Fractional Terms in Sequences

Fractional terms in sequences might seem complex, but they follow an interesting pattern much like the whole numbers do. In this sequence, we see terms like \(\frac{1}{2}, \frac{1}{4}, \frac{1}{6}\), etc. These fractions have a common pattern in their denominators which also forms an arithmetic sequence.

The denominators are 2, 4, 6, ... simply increasing by 2 each time. This means, the denominator of each fractional term can be expressed as \(2n\). Thus, we can understand these as fractions of the form \(\frac{1}{2n}\), where \(n\) denotes the position of the term in the sequence.

• Each fraction's numerator remains constant at 1, making it easier to spot the pattern.
• The denominator follows this simple rule, increasing with every fractional term.

Understanding this pattern allows us to predict any fractional term within a sequence by just knowing its position.

General Term Formula for Mixed Sequences

Finding a general term formula for a sequence that mixes odd whole numbers and fractional terms requires recognizing the pattern switch. In this particular sequence, integer and fractional terms alternate, creating a need for two distinct formulas.

The given sequence indicates:

• If \(n\) is odd, the term is a whole number and follows the formula \(2n - 1\).
• If \(n\) is even, the term is a fraction following the formula \(\frac{1}{2n}\).

This leads us to a combined formula, accommodating both patterns depending on whether \(n\) is odd or even:\[a_n = \begin{cases}2k-1, & \text{if } n = 2k - 1 \\frac{1}{2k}, & \text{if } n = 2k\end{cases}\]Here, \(k\) represents the sequence index indicating whether \(n\) is an odd or even position. This formula helps students easily determine any term in this particular sequence without manually analyzing each term.