Problem 64 - College Algebra with Corequisite Support

Question

List the first five terms of the sequence \(a_{n}=\frac{15 n \cdot(-2)^{n-1}}{47}\),

Step-by-Step Solution

Answer

The first five terms are \(\frac{15}{47}\), \(\frac{-60}{47}\), \(\frac{180}{47}\), \(\frac{-480}{47}\), and \(\frac{1200}{47}\).

Step 1: Identify the formula for the sequence

The given sequence formula is \(a_{n} = \frac{15n \cdot (-2)^{n-1}}{47}\). We need to compute the first five terms, meaning we will use \(n = 1, 2, 3, 4,\) and \(5\).

Step 2: Compute the first term \(a_1\)

Substitute \(n = 1\) into the formula: \(a_1 = \frac{15 \times 1 \cdot (-2)^{1-1}}{47} = \frac{15 \times 1 \cdot 1}{47} = \frac{15}{47}\).

Step 3: Compute the second term \(a_2\)

Substitute \(n = 2\) into the formula: \(a_2 = \frac{15 \times 2 \cdot (-2)^{2-1}}{47} = \frac{15 \times 2 \cdot (-2)}{47} = \frac{-60}{47}\).

Step 4: Compute the third term \(a_3\)

Substitute \(n = 3\) into the formula: \(a_3 = \frac{15 \times 3 \cdot (-2)^{3-1}}{47} = \frac{15 \times 3 \cdot 4}{47} = \frac{180}{47}\).

Step 5: Compute the fourth term \(a_4\)

Substitute \(n = 4\) into the formula: \(a_4 = \frac{15 \times 4 \cdot (-2)^{4-1}}{47} = \frac{15 \times 4 \cdot (-8)}{47} = \frac{-480}{47}\).

Step 6: Compute the fifth term \(a_5\)

Substitute \(n = 5\) into the formula: \(a_5 = \frac{15 \times 5 \cdot (-2)^{5-1}}{47} = \frac{15 \times 5 \cdot 16}{47} = \frac{1200}{47}\).

Key Concepts

Series

In mathematics, a series is the sum of the terms of a sequence. While the exercise focuses on identifying individual terms rather than their sum, understanding series helps deepen comprehension of sequences. A sequence is merely a list of numbers or expressions, typically progressing according to a specific rule or pattern. In contrast, a series takes each term's value and adds them together. This concept lays the groundwork for exploring arithmetic and geometric series.

• Arithmetic series: The difference between consecutive terms is constant. For example, 2 + 4 + 6 + 8 is an arithmetic series where each term increases by 2.
• Geometric series: Each term is a constant multiple of the previous one. For instance, 2 + 4 + 8 + 16 involves each term doubling the last.

Recognizing whether a sequence can form a series relates closely to calculus concepts, such as convergence and divergence. For our original sequence, identifying any potential series might involve exploring successive terms and their behavior. To fully utilize a series-based approach, understanding how each term behaves individually can be crucial for subsequent summation efforts.

Mathematical Formulas

Mathematical formulas are pivotal in finding solutions to sequence-related problems. A formula is essentially a mathematical rule expressed with symbols. It provides a shorthand to calculate terms without manually listing all components repeatedly. In our exercise, the sequence is governed by the formula:\[a_{n} = \frac{15n \cdot (-2)^{n-1}}{47}\]With this, knowing how to substitute values for \(n\), will help in computing individual sequence terms effortlessly.

• Variables: Recognize \(n\) as a variable representing term position in the sequence.
• Substitution: Insert specific values of \(n\) to calculate each term's result. This was used methodically for \(n = 1, 2, 3, 4, 5\) to derive the first five terms.
• Power and Multiplication: Note the term \((-2)^{n-1}\) introduces an alternating pattern through powers, affecting term signs and magnitudes.

Understanding these facets makes calculating and predicting sequence terms much more systematic.

Pattern Recognition

Pattern recognition involves discerning how a series of numbers behaves based on its formula. In the given sequence, we observe specific numerical traits that vary predictably. This skill allows you to anticipate future terms without computing them one by one. To spot patterns in sequences, consider:

• Sign Alternation: The expression \((-2)^{n-1}\) results in negative values for odd \(n\) and positive for even \(n\), creating an alternating sign pattern. Recognizing this supports quick prediction of term nature.
• Magnitude Growth: The linear nature of \(15n\) paired with exponential effects of \((-2)^{n-1}\), shows that apart from sign, terms increase or decrease exponentially in magnitude as \(n\) rises.

Acknowledging these patterns sharpens your ability to efficiently fill gaps in sequences, facilitating a quicker approach to problems potentially carrying hundreds of terms. Analyzing sequence behavior not only aids in current calculations but can reinforce foundational skills applicable to broader mathematical fields.