Problem 16 - Thomas Calculus

Question

In Exercises \(13-26,\) find a formula for the \(n\) th term of the sequence. The sequence \(1,-\frac{1}{4}, \frac{1}{9},-\frac{1}{16}, \frac{1}{25}, \dots\)

Step-by-Step Solution

Answer

\(a_n = \frac{(-1)^{n+1}}{n^2}\) is the formula for the nth term.

Step 1: Identify the Pattern in the Sequence

Examine the sequence given: \(1, -\frac{1}{4}, \frac{1}{9}, -\frac{1}{16}, \frac{1}{25}, \dots\). Notice that the denominators are perfect squares: \(1^2, 2^2, 3^2, 4^2, 5^2, \dots\). This suggests that the denominator of the \(n\)th term is \(n^2\).

Step 2: Determine the Sign Pattern

Observe the signs of the terms in the sequence: the first is positive, the second is negative, the third is positive, and so on. This indicates an alternating pattern of signs. Typically, an alternating pattern can be described by using \((-1)^{n+1}\) to start with a positive sign for the first term.

Step 3: Formulate the General Term of the Sequence

Combine the denominator pattern and sign pattern to form the general term. The nth term of the sequence includes both the alternating sign and the pattern in the denominator: \(a_n = \frac{(-1)^{n+1}}{n^2}\).

Step 4: Verify the Formula with Initial Terms

Check that the formula \(a_n = \frac{(-1)^{n+1}}{n^2}\) gives the correct sequence values. For \(n = 1\), \(a_1 = \frac{(-1)^{2}}{1^2} = \frac{1}{1} = 1\); for \(n = 2\), \(a_2 = \frac{(-1)^{3}}{2^2} = -\frac{1}{4}\), and so on. The formula correctly generates the sequence.

Key Concepts

Alternating Sequence

An alternating sequence is characterized by a pattern of regularly changing signs between positive and negative. This sequence is found in many mathematical contexts as well as real-world applications.

In the provided sequence, you can observe that the terms switch between positive and negative. This is known as an alternating sign pattern. To determine the alternation in this sequence, the formula \((-1)^{n+1}\) is commonly used.

- The expression \((-1)^n\) results in -1 when n is odd and 1 when n is even.- Modifying the exponent to \(n+1\) ensures that the sequence starts with a positive sign when \(n = 1\).

Thus, using \((-1)^{n+1}\) captures the alternating nature by adjusting the sign to align with the sequence’s starting point. This is crucial in defining the general form of the sequence we are tackling.

Denominator Pattern

The denominator in a sequence can often reveal a hidden pattern crucial for formulating the sequence’s general term. In this exercise, the denominators are the perfect squares: 1, 4, 9, 16, 25, etc.

Each of these numbers can be expressed as the square of an integer. Specifically:

• 1 is \(1^2\)
• 4 is \(2^2\)
• 9 is \(3^2\)
• 16 is \(4^2\)
• 25 is \(5^2\)

This sequence suggests a formulaic structure where the denominator of each term is \(n^2\).

When building a general term, recognizing such a denominator pattern simplifies constructing the formula. Understanding this pattern is key, as it directly influences how each term of the sequence is expressed mathematically and confirms the nth term involves the square of n.

General Term of a Sequence

The general term of a sequence provides a formula for finding any term within a sequence efficiently. By combining the patterns observed in the sequence, you arrive at the general term.

In this sequence:

• The sign alternation follows \((-1)^{n+1}\)
• The denominator follows the square pattern \(n^2\)

These two components are integrated to write the general term as \(a_n = \frac{(-1)^{n+1}}{n^2}\).

Using this formula, any term in the sequence can be found without generating all preceding terms, which is highly efficient. It allows calculation of, for instance, the 50th term directly. Understanding the general term crystallizes the essence of the sequence and its behavior, allowing for effective forecasting and analysis of its values.