Problem 54 - Precalculus with Limits

Question

In Exercises 47-62, write an expression for the apparent \(n\)th term of the sequence. (Assume that \( n \) begins with 1.) \( \dfrac{1}{3}, \dfrac{2}{9}, \dfrac{4}{27}, \dfrac{8}{81}, \dots \)

Step-by-Step Solution

Answer

The n-th term of the sequence is given by \(\dfrac{2^{n-1}}{3^n}\).

Step 1: Identify the pattern

The pattern can be observed by looking at how each subsequent term relates to the preceding one. In the given sequence, each term is twice the previous term's numerator and three times the previous term's denominator. Thus, we can conclude that both the numerator and the denominator follow a geometric progression.

Step 2: Express the terms

Let's express the terms of the sequence in the form of a power of the common ratio. The numerator can be expressed as \(2^{n-1}\) (starts with \(2^{1-1} = 1\), and doubles every term) and the denominator as \(3^n\) (starts with \(3^1 = 3\) and triples every term).

Step 3: Formulate the n-th term

Now, combining the expressions for the numerator and the denominator, we can write the n-th term of the sequence as: \(\dfrac{2^{n-1}}{3^n}\). This is the general term for the given sequence.

Key Concepts

Understanding the 'n-th term' in Geometric Sequences

In a geometric sequence, each term is derived from its predecessor by multiplying by a constant called the "common ratio."
The concept of the "n-th term" is invaluable, as it allows us to express a sequence in an elegant and concise way. It refers to the general term of the sequence expressed as a formula depending on the index number, typically represented by "n."
This formula enables prediction of any term's value in the sequence without having to know all the previous terms.
In the given exercise, the sequence \( \frac{1}{3}, \frac{2}{9}, \frac{4}{27}, \frac{8}{81}, \ldots \) is such that each numerator and each denominator follows its own geometric progression.

• The numerators (1, 2, 4, 8, ...) can be represented by the formula: \(2^{n-1}\).
• The denominators (3, 9, 27, 81, ...) can be represented by \(3^n\).

Consequently, this allows us to formulate the n-th term as \( \frac{2^{n-1}}{3^n} \), providing a comprehensive expression to easily compute any term in this sequence.

Recognizing Sequence Patterns

Finding patterns in sequences is all about identifying the rules or relationships that connect each term to the next.
Geometric sequences have a straightforward pattern where you multiply by the same factor each time.
In the given exercise, the pattern can be noticed when observing how each term is developed from the previous one:

• The numerator doubles each step (1 to 2, 2 to 4, 4 to 8, ...).
• The denominator triples each step (3 to 9, 9 to 27, 27 to 81, ...).

Understanding these patterns is crucial, as they allow you to determine how the sequence evolves.
Furthermore, recognizing such patterns can help you derive more general expressions, predict future terms, and even find sums, making pattern recognition a powerful tool in mathematics.

Exploring Progressions in Mathematics

Progressions in mathematics involve sequences of numbers with definitive patterns or relationships.
There are various kinds of progressions, such as arithmetic and geometric, each defined by certain properties.
In geometric progressions, like our sequence in the exercise, every term after the first is derived by multiplying the previous term by a fixed, non-zero number known as the common ratio.
The exercise demonstrates this concept:

• The numerator sequence has a common ratio of 2 (double as you move to the next term).
• The denominator sequence has a common ratio of 3 (triple as you progress).

Understanding how progressions work allows mathematicians to harness them for various applications, from predicting outcomes to solving complex problems in both pure and applied mathematics.