Problem 57
Question
Write an expression for the \(n\) th term of the sequence. (There is more than one correct answer.) \(\frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6}, \ldots\)
Step-by-Step Solution
Verified Answer
The \(n\)th term of the sequence is \(\frac{n+1}{n+2}\).
1Step 1: Identify the Pattern
Observe the terms in the sequence, specifically the numerators and denominators of the fractions. Notice that the numerator and the denominator of each term is \(1\) more than the numerator and the denominator of the preceding term.
2Step 2: Formulate the pattern
Based on the observation in Step 1, the numerator of the \(n\)th term is \(n+1\) and the denominator is \(n+2\). This is because for the first term, the numerator is \(2\) and the denominator is \(3\), and for each subsequent term, the numerator and the denominator increase by \(1\). Thus, the \(n\)th term of the sequence is a fraction where the numerator is \(n+1\) and the denominator is \(n+2\).
3Step 3: Write the \(n\)th term
Using the pattern identified, the \(n\)th term in the sequence can be written as \(\frac{n+1}{n+2}\). This is a general formula that will generate each term of the sequence for any positive integer \(n\).
Key Concepts
Mathematical SequencesFractions in SequencesPattern Recognition in Sequences
Mathematical Sequences
Mathematical sequences are foundations of patterns in numbers, where each term is typically derived from the preceding ones according to a certain rule. It's like a set of dominoes falling one after another, each one connected to the last. In essence, a sequence is a list of numbers that follow a certain order or pattern. Understanding sequences is crucial for various mathematical applications including algebra and calculus.
When it comes to defining sequences, they can either be finite, with a specific number of terms, or infinite, continuing endlessly. One common type of sequence is the arithmetic sequence, where each term after the first is produced by adding a constant to the previous term, like a steady march forward. Another is the geometric sequence, which involves multiplying the preceding term by a constant, resembling an exponential growth. The given exercise deals with a sequence of fractions, where each term is created by applying a consistent alteration to the numerators and denominators of the fractions.
When it comes to defining sequences, they can either be finite, with a specific number of terms, or infinite, continuing endlessly. One common type of sequence is the arithmetic sequence, where each term after the first is produced by adding a constant to the previous term, like a steady march forward. Another is the geometric sequence, which involves multiplying the preceding term by a constant, resembling an exponential growth. The given exercise deals with a sequence of fractions, where each term is created by applying a consistent alteration to the numerators and denominators of the fractions.
Fractions in Sequences
Sequences can involve whole numbers, integers, real numbers, and even complex ones. However, when sequences involve fractions, they bring an additional layer of complexity. Fractions, which consist of numerators and denominators, can create fascinating patterns within sequences.
In the context of the given exercise, each term is a fraction, with both its numerator and denominator following a specific pattern. To decode such patterns, it is important to carefully examine the relationship between the parts of the fraction. Observing that both the numerator and the denominator are increasing by one as we move from one term to the next is essential in discovering the rule that governs the sequence. The exercise provided an excellent example of how recognizing this relationship is used to determine the general nth term of the sequence, which in this case, is written as \(\frac{n+1}{n+2}\).
In the context of the given exercise, each term is a fraction, with both its numerator and denominator following a specific pattern. To decode such patterns, it is important to carefully examine the relationship between the parts of the fraction. Observing that both the numerator and the denominator are increasing by one as we move from one term to the next is essential in discovering the rule that governs the sequence. The exercise provided an excellent example of how recognizing this relationship is used to determine the general nth term of the sequence, which in this case, is written as \(\frac{n+1}{n+2}\).
Pattern Recognition in Sequences
The ability to identify patterns is key to understanding and working with sequences. It involves recognizing a repeatable arrangement or order within the sequence's terms. Pattern recognition is not just about spotting what's next; it's about finding the underlying rule that generates the entire sequence.
In pattern recognition, an effective strategy is to look for changes from one term to the next and to see if those changes carry through consistently. For the given fraction sequence, recognizing that there is a constant increment of one in both the numerator and the denominator from term to term allowed us to derive a general expression for any term within the sequence. Once you've identified this pattern, you can predict any term in the sequence, not just the immediately following one, which is a powerful tool for solving problems involving sequences.
In pattern recognition, an effective strategy is to look for changes from one term to the next and to see if those changes carry through consistently. For the given fraction sequence, recognizing that there is a constant increment of one in both the numerator and the denominator from term to term allowed us to derive a general expression for any term within the sequence. Once you've identified this pattern, you can predict any term in the sequence, not just the immediately following one, which is a powerful tool for solving problems involving sequences.
Other exercises in this chapter
Problem 56
Use the Ratio Test to determine the convergence or divergence of the series. $$ \sum_{n=0}^{\infty} \frac{(-1)^{n} 2^{4 n}}{(2 n+1) !} $$
View solution Problem 56
Find the sum of the series. $$ \sum_{n=0}^{\infty} \frac{(-1)^{n} \pi^{2 n+1}}{3^{2 n+1}(2 n+1) !} $$
View solution Problem 57
Use the Limit Comparison Test to determine the convergence or divergence of the series. $$ \sum_{n=1}^{\infty} \frac{1}{n \sqrt{n^{2}+1}} $$
View solution Problem 57
Comparing Maclaurin Polynomials (a) Compare the Maclaurin polynomials of degree 4 and degree \(5,\) respectively, for the functions \(f(x)=e^{x}\) and \(g(x)=x
View solution