Problem 12 - College Algebra

Question

Write the first four terms of the sequence. $$a_{n}=-4 \cdot(-6)^{n-1}$$

Step-by-Step Solution

Answer

The first four terms are -4, 24, -144, and 864.

Step 1: Identify the expression for the terms

The expression for the sequence given is \( a_{n} = -4 \cdot (-6)^{n-1} \). This will be used to calculate each term by substituting different values of \( n \).

Step 2: Calculate the first term

To find the first term, substitute \( n = 1 \) into the expression: \[a_{1} = -4 \cdot (-6)^{1-1} = -4 \cdot (-6)^{0} = -4 \cdot 1 = -4.\]

Step 3: Calculate the second term

To find the second term, substitute \( n = 2 \) into the expression:\[a_{2} = -4 \cdot (-6)^{2-1} = -4 \cdot (-6)^{1} = -4 \cdot (-6) = 24.\]

Step 4: Calculate the third term

To find the third term, substitute \( n = 3 \) into the expression:\[a_{3} = -4 \cdot (-6)^{3-1} = -4 \cdot (-6)^{2} = -4 \cdot 36 = -144.\]

Step 5: Calculate the fourth term

To find the fourth term, substitute \( n = 4 \) into the expression:\[a_{4} = -4 \cdot (-6)^{4-1} = -4 \cdot (-6)^{3} = -4 \cdot (-216) = 864.\]

Key Concepts

Sequence Terms

In mathematics, a sequence is an ordered list of numbers. Each number in the list is called a term. In the context of a geometric sequence, like the one given in the problem: \[a_{n} = -4 \cdot (-6)^{n-1}\]we refer to \(a_1, a_2, a_3, ...\) as the terms of the sequence. Sequence terms are created by systematically substituting successive integer values for the variable \(n\) into the formula. This approach helps generate each specific term based on its position in the sequence.

• The first term \(a_1\) is obtained by setting \(n = 1\).
• The second term \(a_2\) is obtained by setting \(n = 2\).
• The third term \(a_3\) follows by setting \(n = 3\), and so forth.

This structured approach to how you progress through the sequence ensures consistency, enabling anyone to find any term of the sequence with relative ease.

Exponential Expressions

Exponential expressions are a central part of mathematical sequences, particularly geometric sequences. They involve numbers being raised to powers called exponents. The exponent indicates how many times the base is multiplied by itself. In our sequence, \((-6)^{n-1}\) is an exponential expression where:

• \(-6\) is the base;
• \(n-1\) is the exponent.

This term describes the geometric nature of the sequence. As \(n\) progresses, the evaluation of \((-6)^{n-1}\) causes the sequence to grow or shrink rapidly, depending on the sign and value of the base. Here, the negative base, \(-6\), also introduces an alternating sign behavior.
With every increment in \(n\), the magnitude of the terms often increases significantly, unless modified by a coefficient, such as \(-4\) in this instance, which scales the terms of the sequence.

Substitution Method

The substitution method in mathematics allows you to apply specific values to a general formula to find particular results or terms. In calculating the terms of our sequence, substitution involves plugging in the integer values for \(n\). This is done in a specific order, starting from the smallest positive integer, to successfully determine each subsequent term.To compute a term like \(a_3\):

• Substitute \(n = 3\) into the formula \(a_n = -4 \cdot (-6)^{n-1}\).
• Simplify to calculate \(a_3 = -4 \cdot (-6)^{2} = -4 \cdot 36 = -144\).

This method proves extremely straightforward and systematic, making it excellent for step-wise calculation of sequences.

Recursion Formula

A recursion formula is a type of formula that expresses each term of a sequence in relation to its preceding terms. Although the given sequence does not explicitly use one, understanding recursion helps in diverse mathematical contexts.For example, suppose our sequence was given by a recursion, it might look something like:

• \(a_1 = -4\)
• \(a_{n} = a_{n-1} \cdot (-6)\)

Here, each term is derived by multiplying the previous term by \(-6\). Recursion formulas are useful for efficiently computing terms in sequences where the recursive relationship is easily manageable, especially without recalculations from the base term each time.