Problem 3 - Thomas Calculus

Question

Each of Exercises \(1-6\) gives a formula for the \(n\) th term \(a_{n}\) of a sequence \(\left\\{a_{n}\right\\} .\) Find the values of \(a_{1}, a_{2}, a_{3},\) and \(a_{4} .\) $$ a_{n}=\frac{(-1)^{n+1}}{2 n-1} $$

Step-by-Step Solution

Answer

\(a_1 = 1\), \(a_2 = -\frac{1}{3}\), \(a_3 = \frac{1}{5}\), \(a_4 = -\frac{1}{7}\).

Step 1: Find the First Term

Substitute \(n = 1\) into the formula \(a_n = \frac{(-1)^{n+1}}{2n-1}\). Calculate \(a_1 = \frac{(-1)^{1+1}}{2 \times 1 - 1} = \frac{1}{1} = 1\). Thus, \(a_1 = 1\).

Step 2: Find the Second Term

Substitute \(n = 2\) into the formula. Calculate \(a_2 = \frac{(-1)^{2+1}}{2 \times 2 - 1} = \frac{-1}{3}\). Therefore, \(a_2 = -\frac{1}{3}\).

Step 3: Find the Third Term

Substitute \(n = 3\) into the formula \(a_n\). Compute \(a_3 = \frac{(-1)^{3+1}}{2 \times 3 - 1} = \frac{1}{5}\). Hence, \(a_3 = \frac{1}{5}\).

Step 4: Find the Fourth Term

Substitute \(n = 4\) into the expression \(a_n\). Calculate \(a_4 = \frac{(-1)^{4+1}}{2 \times 4 - 1} = \frac{-1}{7}\). Thus, \(a_4 = -\frac{1}{7}\).

Key Concepts

Understanding the nth Term

The concept of the "nth term" pertains to the general formula that defines the elements of a sequence. In mathematical sequences, each position or term is determined by a specific rule or formula, often involving the variable \(n\). This variable represents the position of a term within the sequence. For any term \(a_n\) in the sequence, \(n\) is a placeholder that can be replaced with any positive integer to find the respective term in the sequence.
For example, given the formula \(a_n = \frac{(-1)^{n+1}}{2n-1}\), it defines a rule where:

• \(n = 1\) gives \(a_1 = 1\)
• \(n = 2\) provides \(a_2 = -\frac{1}{3}\)
• \(n = 3\) results in \(a_3 = \frac{1}{5}\)
• \(n = 4\) produces \(a_4 = -\frac{1}{7}\)

This formula allows us to find any term in the sequence without starting from the first term each time. By substituting any desired term number \(n\), the formula will yield the value of that specific term.

Exploring the Terms of a Sequence

In mathematics, a sequence is essentially a list of numbers arranged in a specific order, with each number referred to as a 'term'. Understanding the terms of a sequence involves recognizing patterns or rules that generate the sequence. Each sequence has its own unique formula that defines how its terms are calculated. By substituting different integer values into the sequence's formula, one can observe how the sequence evolves.
The sequence in question is governed by the formula \(a_n = \frac{(-1)^{n+1}}{2n-1}\), creating a series of terms that change based on the value of \(n\):

• First four terms for \(n = 1\) to \(n = 4\) are \(1, -\frac{1}{3}, \frac{1}{5}, -\frac{1}{7}\)

Each term is distinctly calculated by substituting the respective \(n\) into the formula, revealing a consistent underlying process that defines their relationship. Sequences provide valuable insights into mathematical relationships and can demonstrate both simple and complex patterns.

The Concept of Alternating Series

An alternating series is a particular type of sequence characterized by terms that switch between positive and negative values. This alternation often results from a specific component within the sequence's formula that inherently changes the sign of its terms. Such sequences provide an interesting variation from non-alternating types by showcasing a fluctuating pattern.
In our sequence example, the formula \(a_n = \frac{(-1)^{n+1}}{2n-1}\) is responsible for this alternating behavior:

• The term \((-1)^{n+1}\) is crucial as it causes the sign to switch based on whether \(n\) is odd or even.
• For odd \(n\), the power \(n+1\) results in an even exponent, thus making \((-1)^{n+1} = +1\), rendering the term positive.
• Conversely, for even \(n\), \(n+1\) gives an odd exponent, making \((-1)^{n+1} = -1\), thus ensuring the term is negative.

This type of sequence demonstrates how intricate mathematical formulas can create predictable yet complex behaviors such as oscillating sequences.