Problem 47 - Algebra and Trigonometry Real Mathematics, Real People

Question

Write an expression for the apparent \(n\) th term of the sequence. (Assume \(n\) begins with \(1 .\)) $$\frac{1}{2}, \frac{-1}{4}, \frac{1}{8}, \frac{-1}{16}, \dots$$

Step-by-Step Solution

Answer

The apparent \(n^{th}\) term of the sequence \(\frac{1}{2}, \frac{-1}{4}, \frac{1}{8}, \frac{-1}{16}, \dots\) is \((-1)^{(n+1)} \times \frac{1}{2^n}\).

Step 1: Observing the Pattern

First, observe the sequence and establish the pattern: \(\frac{1}{2}, \frac{-1}{4}, \frac{1}{8}, \frac{-1}{16}, \dots\). After careful observation, we see that all the terms are alternating between positive and negative, starting from positive. In addition, each term is half the magnitude of its predecessor.

Step 2: Formulating the Expression

The expression for the \(n^{th}\) term in the series will be dependent on both the alternating sign and the magnitude of each term. To capture the alternating sign, use \((-1)^{(n+1)}\). For the magnitude, because the sequence is halving each term, use \(\frac{1}{2^n}\). The expression for the apparent \(n^{th}\) term of the series then becomes \((-1)^{(n+1)} \times \frac{1}{2^n}\). This formula multiplies the alternate sign by the divided value.

Step 3: Testing the Expression

Next, it is necessary to test the expression on the sequence to confirm that it is correct. When \(n\) is equal to 1, the expression becomes \((-1)^{(1+1)} \times \frac{1}{2^1} = \frac{1}{2}\) which matches the first term in the sequence. Similarly, for \(n = 2\), it produces \((-1)^{(2+1)} \times \frac{1}{2^2} = \frac{-1}{4}\), the second term in the sequence. It is therefore correct to state that the \(n^{th}\) term of the sequence is \((-1)^{(n+1)} \times \frac{1}{2^n}\).

Key Concepts

Understanding Sequences

A sequence is a set of numbers arranged in a specific order. Each number in this ordered list is known as a 'term'. Sequences can be finite, like a list of elements, or infinite, with numbers continuing indefinitely. The position of each term is denoted by an index, often represented by the letter \(n\). In our given sequence \(\frac{1}{2}, \frac{-1}{4}, \frac{1}{8}, \frac{-1}{16}, \dots\), the terms are organized in a way that they alternate in sign and decrease in magnitude.

• Each term follows a predictable order.
• The sequence begins with a positive number and then alternates between a positive and a negative number.
• The numerator remains constant (either \(1\) or \(-1\)), but the denominator doubles with each subsequent term.

Sequences like this are common in mathematics as they showcase a clear pattern, which can often be expressed as a formula or rule for finding the \(n\)th term. By understanding this concept, we can predict future terms without calculative effort.

Exploring Alternating Series

An alternating series is a sequence of numbers in which the terms alternate in sign. This means that the series can be expressed as a series where positive numbers follow negative numbers, or vice versa. In our sequence, the first term is positive, and each subsequent term switches sign.

• To represent these alternating signs mathematically, we use expressions like \((-1)^{n+1}\), where \(n\) denotes the term's position in the sequence.
• This expression is crucial to ensure that every second term becomes negative, thus alternating the sign from its predecessor.

The use of alternating signs frequently helps in denoting sequences that involve switching states or phases, similar to alternating current in physics. Recognizing an alternating series enables us to form valid expressions that accurately describe the terms within.

Identifying a Mathematical Pattern

In mathematics, a pattern is a predictable and recognizable configuration of numbers, shapes, or other mathematical objects. Our sequence displays a mathematical pattern in its structure:

• The magnitude of each term is precisely half of the one that precedes it, which can be expressed as \(\frac{1}{2^n}\).
• This reduction factor is a common trait in geometric sequences where each term is a constant multiple of the previous one.

Understanding patterns is essential as it allows us to devise formulas, like \((-1)^{(n+1)} \times \frac{1}{2^n}\), to determine any term in the sequence without having to list all previous ones. Patterns make sequences easy to analyze and useful for mathematical modeling in real-world applications.