Problem 32 - Precalculus Mathematics for Calculus

Question

Find the nth 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 \( a_n = 2k - 1 \) for odd \( n \) (\( n = 2k-1 \)), and \( a_n = \frac{1}{2k} \) for even \( n \) (\( n = 2k \)).

Step 1: Understand the Pattern

Examine the given sequence: 1, \( \frac{1}{2} \), 3, \( \frac{1}{4} \), 5, \( \frac{1}{6} \),... Notice that the sequence alternates between whole numbers and fractions.

Step 2: Identify the Whole Number Pattern

The whole numbers appear in positions 1, 3, 5,... This suggests an arithmetic sequence of the form \( a_n = 2n - 1 \), where \( n \) is the term number in this part of the sequence.

Step 3: Identify the Fraction Pattern

The fractions appear in positions 2, 4, 6,..., corresponding to an inverse sequence \( a_n = \frac{1}{2n} \). Notice that the denominator increases by 2 each step.

Step 4: Combine the Patterns

We need to express the whole sequence using the combined pattern. For odd positions \( n = 2k - 1 \), where \( k \) is an integer, use whole numbers. For even positions \( n = 2k \), use fractions. Hence, the nth term of the sequence is: \[a_n = \begin{cases} 2k - 1, & \text{if } n = 2k - 1; \\frac{1}{2k}, & \text{if } n = 2k.\end{cases} \]

Step 5: Verify the Formula

Let's verify for a few terms: - For \( n=1 \), the term is \( a_1 = 2(1) - 1 = 1 \).- For \( n=2 \), the term is \( a_2 = \frac{1}{2(1)} = \frac{1}{2} \).- For \( n=3 \), the term is \( a_3 = 2(2) - 1 = 3 \).These results correspond with the given sequence, confirming that the derived formula is correct.

Key Concepts

Arithmetic Sequence

An arithmetic sequence is a sequence of numbers in which the difference between consecutive terms is constant. This difference is called the common difference.

In the context of our problem, the whole numbers in the sequence (1, 3, 5, ...) follow an arithmetic sequence pattern. Here, each number increases by 2 from the previous one. This suggests that the arithmetic formula is given by \(a_n = 2n - 1\), where \(n\) represents the term number in this part of the sequence.

To understand better, consider:

• The first term \(a_1 = 1\).
• The second term \(a_2 = 3\).
• The third term \(a_3 = 5\).

This confirms the sequence has a common difference of 2, perfectly aligning with the formula \(2n - 1\). Understanding these concepts is crucial when identifying and predicting the terms in an arithmetic sequence.

Alternating Sequences

An alternating sequence is a sequence where the type or nature of numbers changes in a predictable pattern as you move from one element to the next.

In our exercise, the sequence alternates between whole numbers and fractions. Understanding this alternating behavior helps us to determine patterns in the sequence more easily.

The whole numbers appear in every odd position: 1st, 3rd, 5th, etc. Meanwhile, the fractions pop up in every even position: 2nd, 4th, 6th, etc. By recognizing this alternating pattern, we can generate rules for both series and integrate these into the general formula for the entire sequence. The awareness of alternating sequences allows us to dissect complex patterns into manageable parts.

Nth Term

The nth term of a sequence is a formula that allows us to find any term in the sequence without listing all preceding terms.

For the combined sequence in our problem, recognizing the pattern of whole numbers and fractions helped in forming a piecewise formula:
\[ a_n = \begin{cases} 2k - 1, & \text{if } n = 2k - 1; \ \frac{1}{2k}, & \text{if } n = 2k. \end{cases} \]
This formula gives the nth term depending on whether \(n\) is odd or even.

Let's break this down:

• If \(n\) is odd (such as 1, 3, 5,...), it fits into the form \(2k - 1\), meaning these positions produce whole numbers.
• If \(n\) is even (like 2, 4, 6,...), it corresponds to the form \(\frac{1}{2k}\), which defines the fractions with increasing denominators.

With this method, you can determine any term's value directly without manual expansion.

Inverse Sequence

An inverse sequence is where each term is the reciprocal of a number, generally following a specific pattern.

In the context we've been examining, the second sequence in our problem (the fractions) exemplifies this: \( \frac{1}{2}, \frac{1}{4}, \frac{1}{6},...\). Each fraction can be expressed as \( \frac{1}{2k} \) where \(k\) is adjusted to fit the position in the sequence.

The sequence increases with 2 in the denominator stepwise, suggesting the form \( \frac{1}{2n} \), with \(n\) as an index derived from the sequence's structure. Recognizing inverse sequences helps in identifying patterns that aren't immediately arithmetic or geometric, especially in sequences that alternate between different types of numbers.