Problem 16 - Precalculus: Mathematics for Calculus

Question

A sequence is defined recursively by the given formulas. Find the first five terms of the sequence. \(a_{n}=\frac{a_{n-1}}{6}\) and \(a_{1}=-24\)

Step-by-Step Solution

Answer

The first five terms are: \(-24, -4, -\frac{2}{3}, -\frac{1}{9}, -\frac{1}{54}\).

Step 1: Identify the Base Case

The recursive formula provided is \(a_n = \frac{a_{n-1}}{6}\) with the initial condition \(a_1 = -24\). The base case here is the first term of the sequence, \(a_1\). Based on the given information, \(a_1 = -24\).

Step 2: Calculate the Second Term

To find the second term \(a_2\), use the recursive formula. Substitute \(a_1\) in the formula: \(a_2 = \frac{a_1}{6} = \frac{-24}{6} = -4\). Thus, \(a_2 = -4\).

Step 3: Calculate the Third Term

Use the recursive formula to find the third term \(a_3\). Substitute \(a_2\) into the formula: \(a_3 = \frac{a_2}{6} = \frac{-4}{6} = -\frac{2}{3}\). So, \(a_3 = -\frac{2}{3}\).

Step 4: Calculate the Fourth Term

Use the recursive formula to find the fourth term \(a_4\). Substitute \(a_3\) into the formula: \(a_4 = \frac{a_3}{6} = \frac{-\frac{2}{3}}{6} = -\frac{1}{9}\). Therefore, \(a_4 = -\frac{1}{9}\).

Step 5: Calculate the Fifth Term

Use the recursive formula to find the fifth term \(a_5\). Substitute \(a_4\) into the formula: \(a_5 = \frac{a_4}{6} = \frac{-\frac{1}{9}}{6} = -\frac{1}{54}\). Thus, \(a_5 = -\frac{1}{54}\).

Key Concepts

Understanding the Base Case

In recursive sequences, the base case is an essential component. It provides the initial term from which all subsequent terms are calculated. For instance, consider the sequence defined by the formula \(a_n = \frac{a_{n-1}}{6}\) with the base case \(a_1 = -24\). The base case \(a_1\) is crucial because it sets the starting point for the entire sequence. Without this initial term, it would be impossible to determine any further terms.

In simpler words, think of the base case as the "foundation stone" of a building. Just as a sturdy foundation is essential for a stable structure, a well-defined base case is critical for understanding and calculating recursive sequences.

• **Key point**: The base case is your starting term, clearly defined, which can be used to compute the following terms.

Exploring Recursion in Sequences

Recursion refers to the process where a function calls itself in its definition. In the context of sequences, recursion enables the calculation of terms based on preceding terms. This becomes clear with the recursive formula \(a_n = \frac{a_{n-1}}{6}\). Here, each term is derived from the previous one by dividing it by 6.

To comprehend recursion, consider the example of a set of nesting dolls. You start with the largest doll, and to see what's inside, you open it to find a smaller one inside, continuing down in size. Similarly, recursive sequences progressively build each term upon the last.

• The formula \(a_n = \frac{a_{n-1}}{6}\) shows how recursion simplifies the sequence-building process with minimal inputs.
• Important to remember: recursion depends on the base case for initial calculations, each building upon the other like a chain.

Mastering Fraction Division

Fraction division is a pivotal skill in recursive sequences, especially when calculating terms like \(a_3 = \frac{-4}{6} = -\frac{2}{3}\). When you divide fractions, remember the principle: divide by a fraction, multiply by its reciprocal.

For example, to divide fraction \(\frac{-4}{6}\) by 6, transform the operation into multiplication by the reciprocal of 6. The reciprocal of 6 is \(\frac{1}{6}\). So, \(\frac{-4}{6} \times \frac{1}{6} = -\frac{2}{18}\), which simplifies to \(-\frac{1}{9}\).

To tackle fraction division, always:

• Invert (flip) the divisor.
• Multiply the dividend by this reciprocal.
• Simplify the resulting fraction if necessary.

With practice, these steps become second nature in handling recursive sequences that involve fractions. Recursion and fraction division work hand in hand, and mastering both unlocks a better understanding of complex sequences.