Problem 21 - Calculus for Biology and Medicine

Question

Find the next four values of the sequence \(\left\\{a_{n}\right\\}\) on the basis of the values of \(a_{0}, a_{1}, a_{2}, a_{3}, a_{4}\). $$ \frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6} $$

Step-by-Step Solution

Answer

The next four values are \( \frac{6}{7}, \frac{7}{8}, \frac{8}{9}, \frac{9}{10} \).

Step 1: Observe the Pattern

Let's look at the sequence: \( \frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6} \). Notice that each term in the sequence is a fraction, where the numerator and the denominator both increase by 1 for each subsequent term.

Step 2: Identify the Formula

Identify the formula for the sequence terms. The sequence can be represented as \( a_n = \frac{n+1}{n+2} \) for \( n=0, 1, 2, \, ... \).

Step 3: Calculate the Next Term (\(n=5\))

Substitute \( n=5 \) into the formula \( a_5 = \frac{5+1}{5+2} = \frac{6}{7} \).

Step 4: Calculate Another Term (\(n=6\))

Substitute \( n=6 \) into the formula \( a_6 = \frac{6+1}{6+2} = \frac{7}{8} \).

Step 5: Calculate Another Term (\(n=7\))

Substitute \( n=7 \) into the formula \( a_7 = \frac{7+1}{7+2} = \frac{8}{9} \).

Step 6: Calculate the Last Term (\(n=8\))

Substitute \( n=8 \) into the formula \( a_8 = \frac{8+1}{8+2} = \frac{9}{10} \).

Key Concepts

Pattern Observation

When you encounter a mathematical sequence, the first step is to observe the pattern within the sequence. In our exercise, we have a list of fractions:

• \( \frac{1}{2} \)
• \( \frac{2}{3} \)
• \( \frac{3}{4} \)
• \( \frac{4}{5} \)
• \( \frac{5}{6} \)

The key to pattern observation is to look for what's changing as you move from one term to the next. Here, both the numerator and the denominator of each fraction grow by 1 with each subsequent term.
Recognizing this simple addition pattern in both parts of the fractions is crucial to understanding the sequence as a whole. Once you have a clear picture of the increasing pattern, you are well on your way to solving further aspects of the problem. Identifying such patterns not only helps in solving specific sequence problems but also strengthens your ability to recognize and predict numerical patterns in general.

Sequence Formula

Once you have spotted the pattern, the next step is to form a sequence formula that describes the terms. A sequence formula provides a rule that one can use to find any term of the series without having to write out all the preceding terms.
For the given sequence, the formula is \( a_n = \frac{n+1}{n+2} \).

• This means that for any term \(a_n\) , the numerator is one more than the index \(n\), i.e., \(n+1\).
• The denominator is two more than the index \(n\), i.e., \(n+2\).

This formula succinctly captures the rule observed in the pattern. Practicing the development and use of sequence formulas is essential since it allows you to leap directly to finding any specific term in complex sequences, saving time and reducing errors. Understanding the formula gives you the power to not only solve for specific terms but also generalizes your understanding of the sequence's behavior effectively.

Fractional Sequences

Understanding fractional sequences involves recognizing how numerals within fractions change in relation to each other across the sequence. With our sequence, these numerator and denominator relationships are methodical and steadily increasing.
Fractional sequences, like our exercise set, emphasize changes inside a fraction from one term to the next. This often involves considering:

• How the numerator evolves (in our sequence, increment by 1)
• How the denominator evolves (also increment by 1)

Recognizing that fractional sequences follow predictable patterns ensures accurate predictions and calculations of subsequent terms. By mastering fractional sequences, you'll strengthen your ability to handle other varied and complex sequences. This kind of problem allows you to appreciate the beauty of fractions and sequences together, and recognizing structured progressions within them allows you not only to solve such problems but to apply this understanding to a variety of mathematical contexts.