Problem 18 - Contemporary Precalculus

Question

Find a formula for the nth term of the sequence whose first few terms are given. $$\frac{1}{2 \cdot 3}, \frac{1}{3 \cdot 4}, \frac{1}{4 \cdot 5}, \frac{1}{5 \cdot 6}, \frac{1}{6 \cdot 7}, \dots$$

Step-by-Step Solution

Answer

Answer: The general formula for the nth term of the sequence is \(a_n = \frac{1}{n \cdot (n + 1)}\).

Step 1: Identify the pattern

\(\frac{1}{2\cdot3}, \frac{1}{3\cdot4}, \frac{1}{4\cdot5}, \ldots\) The \(n\)-th term has denominator \((n+1)(n+2)\).

Step 2: Write the formula

\(a_n = \frac{1}{(n+1)(n+2)}\)

Key Concepts

Sequence Pattern Recognition

When dealing with sequences, recognizing the pattern is the foundational step in understanding its structure and predicting future terms. For the given sequence \(\frac{1}{2 \cdot 3}, \frac{1}{3 \cdot 4}, \frac{1}{4 \cdot 5}, \frac{1}{5 \cdot 6}, \frac{1}{6 \cdot 7}, \dots\), we observe that each term is the reciprocal of the product of two consecutive integers.

In other words, there’s a clearly defined relationship between the sequence term’s position (usually noted as 'n') and the numbers in the denominator. For the sequence in question, the pattern leads to each term in the denominator being 'n' and (n+1) respectively, where 'n' starts at 2 for the first term. Recognizing this common pattern allows us to formulate a general rule or function that captures the behaviour of the sequence, making it easier to find any term within the sequence.

Consecutive Integers in Sequences

Sequences involving consecutive integers are quite common in precalculus. These integers are numbers listed in order from least to greatest without any gaps. For instance, in the sequence \(\frac{1}{n(n+1)}\), once we identify 'n' as a consecutive integer starting at 2, we can then describe the nth term using these consecutive integers.

Spotting Consecutive Integers

In our example, the denominator consists of two such consecutive numbers ('n' and 'n+1'). This pattern of consecutive integers is not just a characteristic of the given sequence, but it's a common theme across many mathematical contexts, such as algebra and number theory, and understanding it is crucial for solving related problems in mathematics.

Formula Derivation

Deriving a formula for a sequence involves translating the recognized pattern into a mathematical expression that can be used to find any term. After recognizing that our sequence's denominators are the product of consecutive integers, we articulate this relationship as \(a_n = \frac{1}{n(n + 1)}\).

This general formula encompasses the pattern identified and provides a method for calculating any term of the sequence.

Logical Steps in Derivation

To derive the formula, start with the specific examples provided, identify the pattern, and then generalize that pattern to 'n', which represents the position of any term. Deriving such formulas is an essential skill in precalculus studies, aiding in the understanding of series and other complex mathematical concepts.

Precalculus Sequences

Sequences are a fundamental concept in precalculus, which is a branch of mathematics that prepares students for calculus. A sequence is an ordered list of numbers, each of which is defined according to a certain rule or pattern.

The importance of sequences in precalculus lies in their application to various domains like series, limits, and even introduction to functions. Understanding sequence patterns, like the one in our example, serves as a building block to more advanced topics in calculus where sequences evolve into series, and the concept of convergence is introduced. Mastering sequences and their patterns provides a solid foundation for future mathematical concepts that students will encounter.