Problem 30 - Algebra and Trigonometry

Question

\(25-32\) . Find the \(n\) th term of a sequence whose first several terms are given. $$ \frac{3}{4}, \frac{4}{5}, \frac{5}{6}, \frac{6}{7}, \dots $$

Step-by-Step Solution

Answer

The nth term is \(a_n = \frac{n+2}{n+3}\).

Step 1: Identify the Pattern

Start by examining the given terms: \(\frac{3}{4}, \frac{4}{5}, \frac{5}{6}, \frac{6}{7}\). Notice that the numerator and the denominator of each fraction are consecutive integers: \(3, 4, 5, 6, \ldots\) for numerators and \(4, 5, 6, 7, \ldots\) for denominators.

Step 2: Recognize the Sequence Rule

The pattern can be described as: the numerator of the \(n\)-th term is \(n+2\), while the denominator is \(n+3\). We derive this from shifting each position by 2 for the numerator and 3 for the denominator relative to \(n\).

Step 3: Find the General Term

With the pattern identified, formulate the general expression for the \(n\)th term as follows: \(a_n = \frac{n+2}{n+3}\).

Step 4: Verification

Confirm your formula by checking if it works for the initial terms. For \(n=1\), \(a_1 = \frac{1+2}{1+3} = \frac{3}{4}\). For \(n=2\), \(a_2 = \frac{2+2}{2+3} = \frac{4}{5}\), and so forth. This verifies our formula is correct.

Key Concepts

Sequence Patterns

In sequences, patterns help us predict terms even when they aren't listed. Examining sequence patterns lets us understand the relationship between consecutive terms. Consider the sequence given: \( \frac{3}{4}, \frac{4}{5}, \frac{5}{6}, \frac{6}{7}, \ldots \). A pattern arises from the fact that both the numerator and the denominator in each fraction increase by one sequential unit as the sequence progresses.
This particular sequence is called a fractional sequence. Each numerator aligns with a natural number that increases by one as we move forward. Similarly, the denominators follow the same incremental pattern. The predictable nature of these incremental changes facilitates the identification of the overall pattern in the sequence.

• Numerators follow the pattern: 3, 4, 5, 6, ...
• Denominators follow the pattern: 4, 5, 6, 7, ...

This sequence pattern recognition sets the foundation for crafting the formula for any term in the sequence.

Numerator and Denominator

In the sequence, the numerator and denominator play crucial roles in identifying each term. The sequence given earlier showcases how they change from term to term. Each fraction within the sequence has its numerator and denominator increasing by one unit.
To elaborate further:

• For the first term, \( \frac{3}{4} \), the numerator is 3, and the denominator is 4.
• For the second term, \( \frac{4}{5} \), now the numerator has become 4, while the denominator becomes 5.

Understanding the shifts in both numerator and denominator helps in determining the pattern. For any term \( n \):

• The numerator is always \( n + 2 \).
• The denominator is following \( n + 3 \).
  

These expressions represent how any fraction can be constructed in the sequence, essential to formulating a general expression.

General Term Formula

Formulating a general term, or recursive formula, enables us to find any term in the sequence without listing all previous terms. For the given sequence, determining the general formula requires understanding the incremental changes in both numerators and denominators.
Given the patterns observed, we form the general term formula as follows:

• The nth term of the sequence: \( a_n = \frac{n+2}{n+3} \)

This expression allows the computation of any specific term simply by substituting the value of \( n \).
This formula makes it more efficient to find far-off terms without calculating each preceding term, which is vital when sequences grow lengthy.
For example, plug in \( n=1 \) to find that \( a_1 = \frac{1+2}{1+3} = \frac{3}{4} \). Continuing in this manner confirms the formula's reliability.

Mathematical Reasoning

Using mathematical reasoning to confirm that a general term formula matches the given terms is crucial. This process ensures that the formula accurately accounts for all observed patterns in the sequence.
To verify, we match the formula-derived terms against the given sequence terms:

• For \( n=1 \), substitute into the formula: \( a_1 = \frac{3}{4} \), confirming the initial term.
• When \( n=2 \), \( a_2 = \frac{4}{5} \).
• Check further terms in the same way to maintain consistency.

Such verification underpins the importance of mathematical reasoning—using logical thinking to validate our findings. This step ensures that the derived formula holds for all given terms and any potential term in the sequence.
Engaging in such reasoning makes a significant difference in understanding common mathematical practices, ensuring outcomes are both accurate and reliable.