Problem 10 - College Algebra with Modeling and Visualization

Question

Find the first four terms of the sequence. \(a_{n}=(-1)^{n-1}\left(\frac{1}{3^{n}}\right)\)

Step-by-Step Solution

Answer

The first four terms are \(\frac{1}{3}\), \(-\frac{1}{9}\), \(\frac{1}{27}\), and \(-\frac{1}{81}\).

Step 1: Determine the first term

To find the first term of the sequence, substitute \(n=1\) into the sequence formula. \[a_1 = (-1)^{1-1} \left(\frac{1}{3^{1}}\right) = 1 \times \frac{1}{3} = \frac{1}{3}\] Thus, the first term is \(a_1 = \frac{1}{3}\).

Step 2: Determine the second term

Substitute \(n=2\) into the sequence formula to find the second term. \[a_2 = (-1)^{2-1} \left(\frac{1}{3^{2}}\right) = -1 \times \frac{1}{9} = -\frac{1}{9}\] Thus, the second term is \(a_2 = -\frac{1}{9}\).

Step 3: Determine the third term

Substitute \(n=3\) into the sequence formula to find the third term. \[a_3 = (-1)^{3-1} \left(\frac{1}{3^{3}}\right) = 1 \times \frac{1}{27} = \frac{1}{27}\] Thus, the third term is \(a_3 = \frac{1}{27}\).

Step 4: Determine the fourth term

Substitute \(n=4\) into the sequence formula to find the fourth term. \[a_4 = (-1)^{4-1} \left(\frac{1}{3^{4}}\right) = -1 \times \frac{1}{81} = -\frac{1}{81}\] Thus, the fourth term is \(a_4 = -\frac{1}{81}\).

Key Concepts

Geometric Progression

A geometric progression, also called a geometric sequence, is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. In the sequence given by the formula \(a_{n}=(-1)^{n-1}\left(\frac{1}{3^{n}}\right)\), each term involves dividing by increasing powers of 3, hinting at a special kind of geometric progression where the common ratio changes due to the negative factor.Key features of geometric progressions:

• It has a constant ratio between consecutive terms.
• It can involve positive, negative, or fractional common ratios.
• If the common ratio is a fraction less than 1, the terms get smaller.

In this example, the pattern of multiplying by \(-\frac{1}{3}\) leads to decreasing terms as seen in \(-1^{n-1}\) flipping the sign of each next term. Each subsequent term is smaller in absolute value due to the geometric factor \(\frac{1}{3^n}\).

Sequence Formula

The sequence formula is a mathematical expression that defines the pattern of the sequence. In our exercise, the sequence formula is given by:\[a_n = (-1)^{n-1} \left(\frac{1}{3^{n}}\right)\]This tells us how to calculate every term based on its position \(n\) in the sequence. The formula has two key components:

• The power of \(-1\), \((-1)^{n-1}\), alternates the sign with each term.
• The power of \(3\), \(\left(\frac{1}{3^{n}}\right)\), decreases the magnitude of each term by dividing 1 by increasing powers of 3.

This composition of the formula defines how each term changes compared to the previous term, mixing both geometric decrease and alternating signs.

Term Calculation

Calculating terms in sequences involves substituting given \(n\) values into the sequence formula. For the sequence \(a_{n} = (-1)^{n-1}\left(\frac{1}{3^{n}}\right)\), we calculate terms one by one:- **First Term (\(n=1\)):** Substitute \(n=1\) to get \(a_1 = (-1)^{1-1}\left(\frac{1}{3}\right) = \frac{1}{3}\).- **Second Term (\(n=2\)):** Substitute \(n=2\) to get \(a_2 = (-1)^{2-1}\left(\frac{1}{9}\right) = -\frac{1}{9}\).- **Third Term (\(n=3\)):** Substitute \(n=3\) to get \(a_3 = (-1)^{3-1}\left(\frac{1}{27}\right) = \frac{1}{27}\).- **Fourth Term (\(n=4\)):** Substitute \(n=4\) to get \(a_4 = (-1)^{4-1}\left(\frac{1}{81}\right) = -\frac{1}{81}\).Each term involves multiplication and exponents, following the sequence formula precisely.

Alternating Sequence

An alternating sequence switches signs between positive and negative with each successive term. In the given sequence \(a_{n}=(-1)^{n-1}\left(\frac{1}{3^{n}}\right)\), the sign change comes from the \((-1)^{n-1}\) component.Important aspects of alternating sequences:

• Each term switches sign based on its position index \(n\).
• This pattern creates variation, preventing terms from being all positive or negative.
• This alternating behavior is used in many mathematical contexts, like alternating series in calculus.

In our specific example, the alternate sign introduces a dynamic behavior to the sequence, creating a rich pattern as the terms decrease in size while flipping from positive to negative.