Problem 136 - Precalculus with Limits

Question

In Exercises 135-138, find the first five terms of the sequence. \( a_n = \dfrac{(-1)^n x^{2n + 1}}{2n + 1} \)

Step-by-Step Solution

Answer

The first five terms of the sequence defined by \(a_n = \frac{(-1)^n * x^{2n + 1}}{2n + 1}\) are: \(a_1 = -x^3/3\), \(a_2 = x^5/5\), \(a_3 = -x^7/7\), \(a_4 = x^9/9\), and \(a_5 = -x^11/11\).

Step 1: Determine the Pattern for the First Term

First, substitute n = 1 into the definition of the term. The result will be the first term of the sequence: \(a_1 = \frac{(-1)^1 * x^{2*1 + 1}}{2*1 + 1} = -x^3/3\)

Step 2: Determine the Second Term

Substitute n = 2 into the definition. This gives the second term: \(a_2 = \frac{(-1)^2 * x^{2*2 + 1}}{2*2 + 1} = x^5/5\)

Step 3: Find the Third Term

Substitute n = 3 into the equation to get the third term: \(a_3 = \frac{(-1)^3 * x^{2*3 + 1}}{2*3 + 1} = -x^7/7\)

Step 4: Calculate the Fourth Term

By substituting n = 4 into the given equation, we derive the fourth term: \(a_4 = \frac{(-1)^4 * x^{2*4 + 1}}{2*4 + 1} = x^9/9\)

Step 5: Determine the Fifth Term

Lastly, we substitute n = 5. This will give us the fifth term: \(a_5 = \frac{(-1)^5 * x^{2*5 + 1}}{2*5 + 1} = -x^11/11\)

Key Concepts

First Five Terms

When working with sequences, understanding how to find the initial terms is crucial. In this particular problem, you need to find the first five terms of a sequence defined by a specific formula. Let's break this down.

• Start with the formula: \( a_n = \frac{(-1)^n x^{2n + 1}}{2n + 1} \)
• Substitute \( n \) with the first five natural numbers: 1, 2, 3, 4, and 5.
• This gives us the terms:\ \( \begin{align*} a_1 &= \frac{(-1)^1 x^{3}}{3}, \ a_2 &= \frac{x^{5}}{5}, \ a_3 &= \frac{(-1)^3 x^{7}}{7}, \ a_4 &= \frac{x^{9}}{9}, \ a_5 &= \frac{(-1)^5 x^{11}}{11} \end{align*} \)

Identifying these terms helps establish the foundation of the sequence and allows us to better understand its behavior. Each term results from applying the sequence formula, leading to a predictable pattern.

Alternating Series

The given sequence is an excellent example of what is known as an alternating series. An alternating series is characterized by terms that change signs in a regular pattern. In our sequence:

• Odd-positioned terms (\(a_1, a_3, a_5\),...) have a negative sign, because \( (-1)^n \) leads to negative when \( n \) is odd.
• Even-positioned terms (\(a_2, a_4\),...) are positive, as \( (-1)^n \) for even \( n \) results in a positive value.

This alternation is visually represented by the power of \( (-1) \), a common feature in sequences that flip signs. Such series appear frequently in mathematical contexts, notably in harmonic and trigonometric functions.

Sequence Formula

The heart of any sequence lies in its formula. Here, the sequence formula is:
\[ a_n = \frac{(-1)^n x^{2n + 1}}{2n + 1}\]
Each component plays a specific role in shaping the sequence:

• \((-1)^n\): Determines the sign of each term. It switches between -1 and 1, causing the alternation.
• \(x^{2n + 1}\): Decides the power of \( x \) in individual terms, increasing by 2 for subsequent terms.
• \(2n + 1\): The denominator provides the fractional division, affecting term size.

By understanding each part, you can predict terms' properties, like their alternating nature and how they grow larger.

Mathematical Pattern

A key aspect of sequences is recognizing their underlying mathematical pattern. This sequence follows a specific mathematical pattern where:

• The \( n \)-th term structure is consistent—calculate with powers and coefficients based on \( n \).
• Each subsequent term is derived similarly, ensuring predictability.
• Such a pattern simplifies further analysis, such as summing series or checking for convergence.

Understanding patterns in sequences makes it easier to grasp advanced concepts in calculus and algebra. These patterns lend themselves to broader discussions on series expansions, providing a framework for more complex operations.