Problem 18 - Thomas Calculus

Question

In Exercises \(13-26,\) find a formula for the \(n\) th term of the sequence. $$ -\frac{3}{2},-\frac{1}{6}, \frac{1}{12}, \frac{3}{20}, \frac{5}{30}, \dots $$

Step-by-Step Solution

Answer

The formula for the nth term is \(a_n = \frac{-1\times (2n - 5)}{n(n+1)}\).

Step 1: Identify the Pattern

First, observe the sequence to detect any noticeable pattern in the terms given:\(-\frac{3}{2}, -\frac{1}{6}, \frac{1}{12}, \frac{3}{20}, \frac{5}{30}\).Look for any consistent changes between consecutive numerators and denominators.

Step 2: Analyze the Numerators

List the numerators of the sequence terms:\(-3, -1, 1, 3, 5, \dots\).Notice that the numerators appear to increase by 2 as absolute values, if considered without the sign. So, consider the numerators as following the formula:Numerator = \(-1\times (2n - 5)\).

Step 3: Analyze the Denominators

List the denominators of the sequence terms:\(2, 6, 12, 20, 30, \dots\).The differences between the terms are:\(6 - 2 = 4, 12 - 6 = 6, 20 - 12 = 8, 30 - 20 = 10\).This is an arithmetic sequence with a common difference of \(2\). Thus, the sequence can be expressed as:Denominator = \(n(n+1)\).

Step 4: Combine the Patterns

Now combine the formulas for the numerator and the denominator to find the formula of the nth term of the sequence:\(a_n = \frac{-1\times (2n - 5)}{n(n+1)}\).

Step 5: Verify the Formula

Check the formula by substituting small integer values for \(n\) to ensure it generates the terms of the sequence:For \(n = 1\),\(a_1 = \frac{-1\times (2(1) - 5)}{1(1+1)} = \frac{-3}{2}\), matches the first term.For \(n = 2\),\(a_2 = \frac{-1\times (2(2) - 5)}{2(2+1)} = \frac{-1}{6}\), matches the second term.For \(n = 5\),\(a_5 = \frac{-1\times (2(5) - 5)}{5(5+1)} = \frac{5}{30}\), matches the fifth term.This verifies the formula works correctly.

Key Concepts

Arithmetic Sequence

An arithmetic sequence is a sequence of numbers in which the difference between consecutive terms is constant. This constant difference is known as the common difference, and it is crucial for formulating the sequence. Understanding arithmetic sequences can greatly aid in predicting future terms and recognizing patterns in a given sequence.

In the denominator of the original sequence, one can observe an arithmetic sequence. Starting with 2, we see that each subsequent term in the sequence (6, 12, 20, 30,...) seems to grow following a specific pattern. The differences between these terms are 4, 6, 8, and 10, respectively. This shows a consistent addition of 2 to the differences:

• 2 to 6: +4
• 6 to 12: +6
• 12 to 20: +8
• 20 to 30: +10

Identifying an arithmetic sequence in the context of numerators or denominators can assist in developing a formula for the nth term of a more complex sequence, as demonstrated in this exercise.

Numerator and Denominator Analysis

Analyzing both numerators and denominators is key to understanding complex sequences, especially when they differ in behavior. In our sequence, the numerators (-3, -1, 1, 3, 5,...) are quite different from the denominators. However, upon inspection, these numbers increase by 2, when ignoring the negative sign.

For the numerators:

• The absolute changes can be observed as: -3 to -1, -1 to 1, 1 to 3, 3 to 5,...

This reveals a linear progression often expressed with a simple linear formula or expression. Here, the pattern can be captured by the formula: Numerator = \(-1 \times (2n - 5)\). This formula accounts for the increase in value and adjusts for the negative sign required by the terms of the sequence.

On the other hand, the denominators (2, 6, 12, 20,...) follow a distinct pattern.

• The sequence's differences fit the quadratic-like behavior rather than simple arithmetic progression. Observing changes in the sequence, each term matches \(n(n+1)\), suggesting a quadratic formula for the denominators.

Understanding how both numerators and denominators behave independently allows us to construct a formula that accurately depicts complex sequences by harmonizing these two distinct sub-sequences.

nth Term Calculation

Understanding how to calculate the nth term of a sequence is foundational in sequence analysis. It allows for the prediction and calculation of any term without constructing every preceding term, enhancing efficiency and mathematical elegance.

In this sequence, we've identified separate behaviors of the numerator and the denominator. First, by applying their respective formulas, we determine how each component contributes to the sequence's structure:
- Numerator: As discussed, we use \(-1 \times (2n - 5)\), a straightforward formula describing how values change while maintaining the negative nature of every other term.- Denominator: Through deduction, each denominator equates to \(n(n+1)\), incorporating arithmetic progression into a quadratic form.
Once we have these expressions, we combine them to express the nth term formula succinctly:\[a_n = \frac{-1 \times (2n - 5)}{n(n+1)}\]By plugging integer values for \(n\) into this formula, we can verify its accuracy with known terms:

• For \(n = 1\), \(a_1 = \frac{-3}{2}\),
• For \(n = 2\), \(a_2 = \frac{-1}{6}\),
• For \(n = 5\), \(a_5 = \frac{5}{30}\).

This systematic construction and verification reveal how different components of sequences harmonize within the nth term, forming a seamless and complete sequence expression.