Question

Determine the \(n^{\text {th }}\) term of the given sequence. $$3,-\frac{3}{2}, \frac{3}{4},-\frac{3}{8}, \cdots$$

Short answer

The nth term is \(a_n = 3 (-1)^{n+1} \frac{1}{2^{n-1}}\).

Step-by-step solution

Identify the Pattern

Observe the sequence: \(3, -\frac{3}{2}, \frac{3}{4}, -\frac{3}{8}, \cdots\). The terms alternate in sign, suggesting a factor of \((-1)^{n+1}\). The denominators are powers of 2 (1, 2, 4, 8, ...), which signifies a common factor of \(2^{n-1}\). Additionally, the numerator remains constant at 3. Thus, the sequence can be modeled as \(a_n = 3 \times (-1)^{n+1} \times \frac{1}{2^{n-1}}\).

Express the General Term

Combining the observations from Step 1, the general term \(a_n\) can be expressed as: \[ a_n = 3 \times (-1)^{n+1} \times \frac{1}{2^{n-1}} \] This formula represents the nth term of the sequence, accounting for both the alternating sign and the powers of 2 in the denominators.

Key concepts

Nth Term

In sequences, the "nth term" refers to the position of a term within the sequence, enabling us to understand and predict all terms in the sequence based on their position. When examining a sequence, identifying the nth term gives a formulaic approach to find any term without needing to list out the entire sequence. For example, in the sequence given in our exercise, each position or term can be defined by the formula: \[ a_n = 3 imes (-1)^{n+1} imes \frac{1}{2^{n-1}} \].This formula reveals how each term in the sequence relates to its position \( n \), and allows you to calculate the value of any nth term directly.

Alternating Series

An alternating series is a series of numbers in which the signs of the terms switch back and forth. This is characterized by the presence of terms that have different signs, typically alternating between positive and negative. This is usually represented by the factor \((-1)^{n}\) or \((-1)^{n+1}\), where the exponent \(n\) helps determine whether a term is positive or negative. In our sequence: \(3, -\frac{3}{2}, \frac{3}{4}, -\frac{3}{8}, \cdots\), the negative and positive terms alternate due to the \((-1)^{n+1}\) factor. This is critical in providing the alternating sign, which produces a series that switches between adding and subtracting each subsequent term in the sequence.

Powers of Two

A power of two is a number that can be expressed as \(2^n\), where \(n\) is a non-negative integer. Powers of two are foundational in sequences, especially those involving fractions, as denominators or factors that multiply terms. In the example from the exercise, the denominators of each term are powers of two: 1, 2, 4, 8, and so on.The sequence's terms become fractions with these powers appearing in the denominator. This determines the size of each fraction relative to others. The formula \(\frac{1}{2^{n-1}}\) in our sequence denotes the presence of powers of two, which help adjust the value of each term by shrinking it according to its position \(n\). Every time \(n\) increases, the denominator, being a power of two, makes the term smaller.

General Term Formula

The general term formula is a mathematical expression that provides a way to calculate any term within a sequence based on its position. Rather than manually calculating each term, this formula simplifies the process, specifying how each term is constructed and how different factors contribute to its value.For our given sequence, the general term formula \[ a_n = 3 \times (-1)^{n+1} \times \frac{1}{2^{n-1}} \] describes the alternating sign, constant numerator, and a fractional denominator powered by two.

• The constant multiplier "3" is the common factor throughout, seen in all terms of the sequence.
• The factor \((-1)^{n+1}\) ensures the sequence terms alternate between positive and negative signs.
• The fraction \(\frac{1}{2^{n-1}}\) highlights the shrinking effect through powers of two as \(n\) increases.

This general term formula, therefore, encapsulates the entire behavior of the sequence, making it easy to determine any specific term.