Problem 16 - Thomas Calculus

Question

Find a formula for the \(n\)th term of the sequence. $$ 1,-\frac{1}{4}, \frac{1}{9},-\frac{1}{16}, \frac{1}{25}, \ldots $$

Step-by-Step Solution

Answer

The \(n\)th term of the sequence is \(a_n = \frac{(-1)^{n+1}}{n^2}\).

Step 1: Identify the Pattern in the Sequence

Examine the given sequence: \(1, -\frac{1}{4}, \frac{1}{9}, -\frac{1}{16}, \frac{1}{25}, \ldots\). Observe that the numerators of the fractions are alternating between positive and negative signs. Also, note that the denominators are perfect squares: \(1^2, 2^2, 3^2, 4^2,\) and so on.

Step 2: Generalize the Denominator

Recognize that the denominator of each term is the square of the term's position in the sequence. For example, the first term has a denominator of \(1^2\), the second term has \(2^2\), the third term has \(3^2\), etc. Thus, the denominator for the \(n\)th term will be \(n^2\).

Step 3: Determine the Numerator and Sign

The numerator alternates in sign: \(1, -1, 1, -1,\ldots\). Notice that this can be represented by the expression \((-1)^{n+1}\). The pattern stems from powers of \(-1\), where odd powers give positive \(1\) and even powers give negative \(1\). As such, each numerator is \((-1)^{n+1}\).

Step 4: Formulate the General Term Formula

Combine the generalized numerator and denominator to form the \(n\)th term of the sequence. The \(n\)th term formula is \(a_n = \frac{(-1)^{n+1}}{n^2}\). This formula captures both the alternating signs and the pattern of the denominators.

Key Concepts

Alternating Signs

An essential feature of the sequence is the alternating signs in the numerators. Alternating signs mean that the sequence changes from positive to negative in a regular pattern. This can be observed in sequences like the given one:

• 1
• -1/4
• 1/9
• -1/16
• 1/25

The numerators switch signs between each consecutive term, giving us a wave of ups and downs. This is mathematically expressed using powers of -1. For the sequence above, this pattern arises from the expression \((-1)^{n+1}\), where:

• Odd values of \(n\) yield a positive result.
• Even values of \(n\) yield a negative result.

Understanding this alternating sign pattern is crucial for predicting the behavior of future terms.

Perfect Squares

Observe the denominators in the sequence. Every term's denominator is a perfect square, reflecting an important number pattern. Perfect squares are numbers that are the square of an integer. In our sequence, these are:

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

Each denominator fits the form \(n^2\), where \(n\) corresponds to the position of the term in the sequence. Recognizing perfect squares helps us to generalize the formula for the sequence and is key in identifying the sequence's progression.

Numerator and Denominator

Each term in the sequence can be split into two distinct parts: the numerator and the denominator. Here’s how they work together:

• **Numerator**: This part involves alternating between positive and negative and can be represented using \((-1)^{n+1}\), leveraging the power properties of -1.
• **Denominator**: The position of each term determines this part, and it's a perfect square \(n^2\) for each term position \(n\).

Understanding both components is essential for writing a complete sequence formula. The numerator accounts for the sign, while the denominator gives the position-based value.

General Term Formula

Combining all the observed patterns of the sequence into one equation yields the general term formula. This is crucial to predicting any term in the sequence without listing them out individually. The general term formula for our sequence is:\[a_n = \frac{(-1)^{n+1}}{n^2}\]Here’s how it breaks down:

• The expression \((-1)^{n+1}\) gives the correct sign for any term, alternating between positive and negative.
• The denominator \(n^2\) provides the structure of increasing perfect squares as the sequence advances.

This concise formula efficiently encodes the alternating signs and perfect square patterns, making it quick to calculate any term’s value in the sequence.