Problem 31 - Precalculus with Limits

Question

In Exercises 9-32, write the first five terms of the sequence. (Assume that \( n \) begins with 1.) \( a_n = \dfrac{(-1)^{n + 1}}{n^2 + 1} \)

Step-by-Step Solution

Answer

The first five terms of the sequence are \( \frac{1}{2}, -\frac{1}{5}, \frac{1}{10}, -\frac{1}{17}, \frac{1}{26} \).

Step 1: Identify the first term

Substitute n = 1 into the formula: \( a_1 = \dfrac{(-1)^{1 + 1}}{1^2 + 1} = \dfrac{(-1)^2}{1 + 1} = \dfrac{1}{2}\).

Step 2: Identify the second term

Substitute n = 2 into the formula: \( a_2 = \dfrac{(-1)^{2 + 1}}{2^2 + 1} = \dfrac{-1}{4 + 1} = -\dfrac{1}{5}\).

Step 3: Identify the third term

Substitute n = 3 into the formula: \( a_3 = \dfrac{(-1)^{3 + 1}}{3^2 + 1} = \dfrac{1}{9 + 1} = \dfrac{1}{10}\).

Step 4: Identify the fourth term

Substitute n = 4 into the formula: \( a_4 = \dfrac{(-1)^{4 + 1}}{4^2 + 1} = \dfrac{-1}{16 + 1} = -\dfrac{1}{17}\).

Step 5: Identify the fifth term

Substitute n = 5 into the formula: \( a_5 = \dfrac{(-1)^{5 + 1}}{5^2 + 1} = \dfrac{1}{25 + 1} = \dfrac{1}{26}\).

Key Concepts

Sequence Terms

A sequence is an ordered list of numbers that follows a certain pattern or rule. In the exercise provided, the rule for generating the sequence terms is given by the formula \( a_n = \dfrac{(-1)^{n + 1}}{n^2 + 1} \). This formula determines the value of each term based on its position in the sequence, indexed by \( n \).
To find the sequence terms, you substitute different values of \( n \) into the formula. For example, when \( n = 1 \), the term is \( a_1 = \dfrac{1}{2} \), and similarly for other values of \( n \).
The important thing to note here is the behavior of the sequence as \( n \) increases. Each term will be calculated uniquely based on this formula, creating a distinct pattern in the sequence.
By understanding how each term is generated, you can predict further terms and grasp the overall structure of the sequence more clearly.

Alternating Signs

Alternating signs in a sequence refer to the change between positive and negative terms as you progress through the sequence. This is controlled by the \((-1)^{n+1}\) part of the formula \( a_n = \dfrac{(-1)^{n + 1}}{n^2 + 1} \).
Here's how it works:

• For odd \( n \), \( n+1 \) is even, making \((-1)^{n+1} = 1\) resulting in a positive term.
• For even \( n \), \( n+1 \) is odd, making \((-1)^{n+1} = -1\) resulting in a negative term.

Therefore, this pattern causes the sequence to alternate signs between each term.
In our sequence, the first term \( a_1 \) is positive, the second term \( a_2 \) is negative, the third term \( a_3 \) becomes positive again, and so on. Understanding this aspect is crucial for predicting and analyzing the behavior of the sequence as it progresses.

Formula Manipulation

Formula manipulation involves substituting values into a formula to derive specific outcomes, in this case, sequence terms. Let's break it down with the sequence formula \( a_n = \dfrac{(-1)^{n + 1}}{n^2 + 1} \).
To manipulate this formula, you start by replacing \( n \) with any integer to calculate the respective term. For example:

• For \( n = 1 \), substitute into the formula, yielding \( a_1 = \dfrac{(-1)^2}{1 + 1} = \dfrac{1}{2} \).
• Similarly, substitute \( n = 2 \) to get \( a_2 = -\dfrac{1}{5} \), and so on for other values of \( n \).

This method also requires simplification skills, such as calculating \((-1)^{n+1}\) and squaring \( n \), which play crucial roles in determining the sign and denominator of each term.
Understanding how to manipulate the formula effectively allows you to calculate any term in the sequence confidently. This practice reinforces your skills in mathematical substitution and simplification.