Problem 17 - Calculus: Early Transcendentals

Question

Find a formula for the general term \( a_n \) of the sequence, assuming that the pattern of the first few terms continues. \( \left\\{\begin{array} \frac {1}{2}, - \frac {4}{3}, \frac {9}{4}, - \frac {16}{5}, \frac {25}{6}, . . . . .\end{array}\right\\} \)

Step-by-Step Solution

Answer

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

Step 1: Recognize the Pattern in the Sequence

First, observe the sequence provided: \( \frac{1}{2}, -\frac{4}{3}, \frac{9}{4}, -\frac{16}{5}, \frac{25}{6} \). Note that the numerators are \( 1, 4, 9, 16, 25 \) which are perfect squares: \( 1^2, 2^2, 3^2, 4^2, 5^2 \). The denominators are consecutive integers starting from 2.

Step 2: Extract the General Formula for the Numerator

Identify that each numerator follows \( n^2 \) where \( n \) is the term number. So, for the nth term, the numerator is \( n^2 \).

Step 3: Extract the General Formula for the Denominator

Identify the pattern in denominators, which starts from 2 and increases by 1 for each consecutive term. Therefore, the denominator is \( n+1 \).

Step 4: Determine the Sign Pattern of the Sequence

Observe the signs of the terms: positive, negative, positive, negative, positive. This indicates an alternating sign pattern: \(( -1 )^{n+1}\) alternates the signs starting positive for \( n = 1 \).

Step 5: Combine the Results to Formulate the General Term

By combining these components, we can form a formula for the general term of the sequence: \[ a_n = \frac{(-1)^{n+1} \cdot n^2}{n+1} \]

Key Concepts

Numerator Pattern

The numerators in the given sequence follow a simple and clear pattern. If you list them, you get: 1, 4, 9, 16, 25.
Each of these numbers is a perfect square. These are the squares of the sequence 1, 2, 3, 4, 5, and so on.
This means that for the nth term, the numerator is given by the formula \( n^2 \).

Understanding this numerator pattern is crucial because it repeats uniformly and helps predict further terms in the sequence.

• * First term: 1 is \( 1^2 \)
• * Second term: 4 is \( 2^2 \)
• * Third term: 9 is \( 3^2 \)
• * Fourth term: 16 is \( 4^2 \)
• * Fifth term: 25 is \( 5^2 \)

This predictable pattern allows us to easily define any numerator in the sequence using the simple square relation.

Denominator Pattern

The denominators of the sequence also display a straightforward pattern. Take a look at the denominators: 2, 3, 4, 5, 6.
These are simply consecutive integers starting from 2.
This means, to find the denominator for the nth term, you simply add 1 to the term number. Hence, the formula for any denominator is \( n+1 \).

• * First term denominator: 2 (i.e., 1+1)
• * Second term denominator: 3 (i.e., 2+1)
• * Third term denominator: 4 (i.e., 3+1)
• * Fourth term denominator: 5 (i.e., 4+1)
• * Fifth term denominator: 6 (i.e., 5+1)

This pattern is consistent and helps us determine the denominator for any term further in the sequence.

Alternating Sequence

The alternating nature of the sequence is captured vividly in the sign change between consecutive terms.
If we check the sequence: \( \frac{1}{2} \), \(-\frac{4}{3} \), \( \frac{9}{4} \), \(-\frac{16}{5} \), \( \frac{25}{6} \), we notice a regular flip in signs:

• The first term is positive.
• The second term is negative.
• The third term is positive, and so on.

This pattern follows an alternating sequence, specifically represented by \( (-1)^{n+1} \).
This expression alternates the sign for each term, starting with positive on the first term when \( n = 1 \).
Such alternating sequences are common and essential, especially in calculus and series, as they help define the shift between succeeding terms.

General Term Derivation

Once all patterns are identified, we can derive the general term formula for the sequence.
By combining the numerator, denominator, and alternating sign components, the general formula becomes:
\[ a_n = \frac{(-1)^{n+1} \cdot n^2}{n+1} \] This expression captures the nature of the sequence accurately:

• The numerator \( n^2 \) comes from the perfect square sequence.
• The denominator \( n+1 \) is due to consecutive integers starting from 2.
• The sign change is modeled by \( (-1)^{n+1} \).

This formula provides a quick and efficient way to determine any term in the sequence, allowing for both predictions and further analysis in mathematical contexts like series evaluations or pattern recognitions.