Problem 29 - Algebra and Trigonometry

Question

\(25-32\) . Find the \(n\) th term of a sequence whose first several terms are given. $$ 1, \frac{3}{4}, \frac{5}{9}, \frac{7}{16}, \frac{9}{25}, \dots $$

Step-by-Step Solution

Answer

The \( n \)th term is \( \frac{2n-1}{n^2} \).

Step 1: Identify the Numerator Pattern

First, we should analyze the sequence of numerators: 1, 3, 5, 7, 9. This sequence increases by 2 with each successive term. Hence, this is an arithmetic sequence with a common difference of 2. The general form of the numerator for the \( n \)th term is: \( 2n - 1 \).

Step 2: Identify the Denominator Pattern

Next, analyze the sequence of denominators: 1, 4, 9, 16, 25. Notice that these are perfect squares: \( 1^2, 2^2, 3^2, 4^2, 5^2 \). Thus, the denominator for the \( n \)th term can be represented as \( n^2 \).

Step 3: Formulate the General Term

Combine the patterns identified in the numerators and denominators. The \( n \)th term of the sequence is the ratio of their respective arithmetic and geometric terms: \( \frac{2n-1}{n^2} \).

Step 4: Verify the Formula With Given Terms

To ensure accuracy, substitute \( n = 1, 2, 3, 4, 5 \) into the formula \( \frac{2n-1}{n^2} \):- For \( n = 1: \frac{2(1)-1}{1^2} = \frac{1}{1} = 1 \)- For \( n = 2: \frac{2(2)-1}{2^2} = \frac{3}{4} \)- For \( n = 3: \frac{2(3)-1}{3^2} = \frac{5}{9} \)- For \( n = 4: \frac{2(4)-1}{4^2} = \frac{7}{16} \)- For \( n = 5: \frac{2(5)-1}{5^2} = \frac{9}{25} \)The calculations confirm that the formula is correct.

Key Concepts

Arithmetic Sequence

An arithmetic sequence is a series of numbers in which each term after the first is obtained by adding a constant. This constant is known as the common difference. In the given numerator sequence: 1, 3, 5, 7, 9, the common difference is 2. Each number is 2 greater than the one before it.
To find any term in an arithmetic sequence, we use the formula:

• \( a_n = a_1 + (n-1) imes d \)

where:

• \( a_1 \) is the first term in the sequence, and here it is 1.
• \( d \) is the common difference, which is 2 in this example.
• \( n \) is the position of the term in the sequence.

Using this formula, the numerators for any \( n \)th term can be determined as \( 2n - 1 \). This formula is derived by substituting \( a_1 = 1 \) and \( d = 2 \).

Geometric Sequence

A geometric sequence involves a series of numbers in which each term after the first is obtained by multiplying the previous term by a fixed, non-zero number called the common ratio. However, in the context of the original problem, the focus is on the pattern involved with the denominators: 1, 4, 9, 16, 25.
Interestingly, these numbers are not part of a geometric sequence. They're actually perfect squares of consecutive integers: \( 1^2, 2^2, 3^2, 4^2, 5^2 \).

For this sequence:

• The \( n \)th term of the denominator is given as \( n^2 \).

Understanding this pattern is crucial since it helps in defining the denominator part of the formula for the sequence. Essentially, each denominator is derived by squaring the position number \( n \).

Nth Term Formula

The nth term formula for a sequence is a mathematical expression that allows us to calculate any term in the sequence systematically. By identifying and combining patterns from both the numerators and denominators, we arrive at the comprehensive nth term formula.
Let's combine our observations from both sequences:

• The numerator follows an arithmetic sequence formula: \( 2n - 1 \).
• The denominator follows a perfect square pattern: \( n^2 \).

By representing the nth term as a fraction of the numerator over the denominator, we establish the complete formula for any term in the sequence:

• \( a_n = \frac{2n - 1}{n^2} \)

To ensure its correctness, substitute various values of \( n \) (such as 1, 2, 3, 4, and 5) into the formula. Doing this confirms it produces the correct sequence values listed in the original problem. This powerful formula allows predicting any term in the sequence without needing to list all prior terms.