Problem 30
Question
Find a formula for the \(n\)th term of the sequence. $$ \sqrt{\frac{5}{8}}, \sqrt{\frac{7}{11}}, \sqrt{\frac{9}{14}}, \sqrt{\frac{11}{17}}, \dots $$
Step-by-Step Solution
Verified Answer
\( a_n = \sqrt{\frac{2n + 3}{3n + 5}} \)
1Step 1: Identify the Pattern in the Numerator
Observe the numerators of the sequence terms: 5, 7, 9, 11. We notice that the numerators form an arithmetic sequence with the first term 5 and a common difference of 2. Thus, the general formula for the nth term of the numerators is: \( 5 + 2(n-1) = 2n + 3 \).
2Step 2: Identify the Pattern in the Denominator
Examine the denominators of the sequence terms: 8, 11, 14, 17. The denominators also form an arithmetic sequence with the first term 8 and a common difference of 3. Therefore, the formula for the nth term of the denominators is: \( 8 + 3(n-1) = 3n + 5 \).
3Step 3: Formulate the General Term for the Sequence
The general term of the sequence can be expressed as a fraction formed by the formulas derived in Step 1 and Step 2. Therefore, the nth term of the sequence is: \( \sqrt{\frac{2n + 3}{3n + 5}} \).
Key Concepts
Arithmetic SequenceNumerator PatternDenominator PatternGeneral TermNth Term Formula
Arithmetic Sequence
An arithmetic sequence is a sequence of numbers where each term after the first is derived by adding a constant, called the "common difference," to the previous term.
This common difference remains the same throughout the entire sequence.
In the given exercise, both the numerators \(5, 7, 9, 11\) and denominators \(8, 11, 14, 17\) of the fractions form their own arithmetic sequences.
This common difference remains the same throughout the entire sequence.
In the given exercise, both the numerators \(5, 7, 9, 11\) and denominators \(8, 11, 14, 17\) of the fractions form their own arithmetic sequences.
- The numerators start at 5 and increase by 2. Thus, they form the sequence: 5, 7, 9, 11,...
- The denominators start at 8 and increase by 3. Thus, they form the sequence: 8, 11, 14, 17,...
Numerator Pattern
The numerator pattern involves identifying which arithmetic sequence the numerators \(5, 7, 9, 11, \ldots\) belong to.
To find a formula for this pattern, observe that the numerators increase by 2 with each successive term.This pattern means:- The first term \(a_1\) is 5.- The common difference \(d\) is 2.Using the arithmetic sequence formula for the nth term \(a_n = a_1 + (n-1) \cdot d\), we get:\[ a_n = 5 + (n-1) \cdot 2 = 2n + 3 \]This formula \(2n + 3\) enables us to find the numerator for the nth term easily.
To find a formula for this pattern, observe that the numerators increase by 2 with each successive term.This pattern means:- The first term \(a_1\) is 5.- The common difference \(d\) is 2.Using the arithmetic sequence formula for the nth term \(a_n = a_1 + (n-1) \cdot d\), we get:\[ a_n = 5 + (n-1) \cdot 2 = 2n + 3 \]This formula \(2n + 3\) enables us to find the numerator for the nth term easily.
Denominator Pattern
The denominator pattern for the sequence \(8, 11, 14, 17, \ldots\) follows a similar process to the numerators.
Here, the denominators form an arithmetic sequence with a different pattern.In this case:- The first term \(b_1\) is 8.- The common difference \(d\) is 3.By applying the arithmetic sequence formula for the nth term \(b_n = b_1 + (n-1) \cdot d\), we determine:\[ b_n = 8 + (n-1) \cdot 3 = 3n + 5 \]This formula \(3n + 5\) allows for determining the denominator of the nth term efficiently.
Here, the denominators form an arithmetic sequence with a different pattern.In this case:- The first term \(b_1\) is 8.- The common difference \(d\) is 3.By applying the arithmetic sequence formula for the nth term \(b_n = b_1 + (n-1) \cdot d\), we determine:\[ b_n = 8 + (n-1) \cdot 3 = 3n + 5 \]This formula \(3n + 5\) allows for determining the denominator of the nth term efficiently.
General Term
The general term of a sequence integrates both the numerator and the denominator formulas.
This forms the complete fraction for any nth term. In the provided sequence, each term is represented as a square root of a fraction.By combining formulas:- Numerator formula: \(2n + 3\)- Denominator formula: \(3n + 5\)The general term, \(T_n\), is constructed as:\[ T_n = \sqrt{\frac{2n + 3}{3n + 5}} \]This expression simplifies the process of finding any term in the sequence, allowing students to calculate the nth term directly without listing all preceding terms.
This forms the complete fraction for any nth term. In the provided sequence, each term is represented as a square root of a fraction.By combining formulas:- Numerator formula: \(2n + 3\)- Denominator formula: \(3n + 5\)The general term, \(T_n\), is constructed as:\[ T_n = \sqrt{\frac{2n + 3}{3n + 5}} \]This expression simplifies the process of finding any term in the sequence, allowing students to calculate the nth term directly without listing all preceding terms.
Nth Term Formula
The nth term formula is a specific expression that defines any term in a sequence based on its position \(n\).
This formula is particularly useful as it provides a direct way to find the value of a term without generating all previous terms.For this exercise:- The term is formed as \(T_n = \sqrt{\frac{2n + 3}{3n + 5}}\).This nth term formula is derived by combining separate sequences seen in the numerators and denominators.
The beauty of the nth term formula is that it applies universally within the defined sequence, ensuring every term calculated adheres to the same structural pattern. Thus, leveraging these mathematical expressions not only simplifies computations but also deepens the understanding of patterns within sequences.
This formula is particularly useful as it provides a direct way to find the value of a term without generating all previous terms.For this exercise:- The term is formed as \(T_n = \sqrt{\frac{2n + 3}{3n + 5}}\).This nth term formula is derived by combining separate sequences seen in the numerators and denominators.
The beauty of the nth term formula is that it applies universally within the defined sequence, ensuring every term calculated adheres to the same structural pattern. Thus, leveraging these mathematical expressions not only simplifies computations but also deepens the understanding of patterns within sequences.
Other exercises in this chapter
Problem 30
In Exercises \(17-46,\) use any method to determine if the series converges or diverges. Give reasons for your answer. $$ \sum_{n=1}^{\infty}\left(\frac{1}{n}-\
View solution Problem 30
Express each of the numbers in Exercises \(23-30\) as the ratio of two integers. $$3 . \overline{142857}=3.142857142857 \ldots$$
View solution Problem 31
Find the Taylor series generated by \(f\) at \(x=a.\) \(f(x)=e^{x}, \quad a=2\)
View solution Problem 31
Which of the series in Exercises 13 46 converge, and which diverge? Give reasons for your answers. (When you check an answer, remember that there may be more th
View solution