Problem 25 - Precalculus with Limits

Question

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

Step-by-Step Solution

Answer

The first five terms of the sequence are -1, \frac{1}{4}, -\frac{1}{9}, \frac{1}{16}, -\frac{1}{25}.

Step 1 - Identifying the Formula

The formula given to generate the sequence is \(a_n = \frac{(-1)^n}{n^2}\). This will generate the nth term of the sequence when values of n are substituted into the expression.

Step 2 - Substitute n=1

Let's substitute n=1 into the sequence formula: \(a_1 = \frac{(-1)^1}{1^2}\). Simplifying this gives \(a_1 = -1\). So, the first term of the sequence is -1.

Step 3 - Substitute n=2

Now let's substitute n=2 into the sequence formula: \(a_2 = \frac{(-1)^2}{2^2}\). Simplifying this gives \(a_2 = \frac{1}{4}\). So, the second term of the sequence is \frac{1}{4}.

Step 4 - Substitute n=3

Next, substitute n=3 into the sequence formula: \(a_3 = \frac{(-1)^3}{3^2}\). This simplifies to \(a_3 = -\frac{1}{9}\). The third term of the sequence is -\frac{1}{9}.

Step 5 - Substitute n=4

Next, substitute n=4 into the sequence formula: \(a_4 = \frac{(-1)^4}{4^2}\). Simplifying gives \(a_4 = \frac{1}{16}\). So, the fourth term of the sequence is \frac{1}{16}.

Step 6 - Substitute n=5

Finally, substitute n=5 into the sequence formula: \(a_5 = \frac{(-1)^5}{5^2}\). This simplifies to \(a_5 = -\frac{1}{25}\). The fifth term of the sequence is -\frac{1}{25}.

Key Concepts

Alternating Sequences

In mathematics, an alternating sequence is a sequence where the signs of the terms change from positive to negative or vice versa as the sequence progresses. This usually results in a pattern such as positive, negative, positive, negative, and so forth. It creates a zig-zag pattern when graphed.
The signs of the sequence values may be determined by a factor like \((-1)^n\) in the sequence formula, as seen in the original exercise.

• If \(n\) is odd, \((-1)^n\) results in -1, making the term negative.
• If \(n\) is even, \((-1)^n\) results in 1, making the term positive.

This alternation adds an interesting behavior to sequences, influencing how series sum up and how convergence is studied.

Sequence Formula

The sequence formula defines how the terms in a sequence are generated. It is essentially an equation that symbolizes the nth term of a sequence.
In the exercise, the sequence formula is \(a_n = \frac{(-1)^n}{n^2}\). This formula uses division and powers to change the magnitudes and signs of each term.

• The factor \((-1)^n\) decides the sign of each term, alternating between positive and negative.
• The denominator \(n^2\) indicates that as \(n\) increases, the term's magnitude decreases since the square of \(n\) grows.

Knowing how to interpret and apply the formula is crucial, as it guides in determining the characteristics and behavior of any given sequence.

Sequence Terms

Sequence terms are the individual components that make up a sequence. Each term is calculated using the sequence formula by substituting the respective value of \(n\). In the exercise, the first few terms were calculated by individually plugging in values from \(n=1\) to \(n=5\).

• Term 1: \(a_1 = \frac{(-1)^1}{1^2} = -1\)
• Term 2: \(a_2 = \frac{(-1)^2}{2^2} = \frac{1}{4}\)
• Term 3: \(a_3 = \frac{(-1)^3}{3^2} = -\frac{1}{9}\)
• Term 4: \(a_4 = \frac{(-1)^4}{4^2} = \frac{1}{16}\)
• Term 5: \(a_5 = \frac{(-1)^5}{5^2} = -\frac{1}{25}\)

These calculations demonstrate how the sequence formula operates, and how alternatively the signs and diminishing sizes of terms are structured. This is crucial for understanding how terms in a sequence behave and interact with each other.