Problem 6 - Essential Calculus Early Transcendentals

Question

\(5-8=\) Find a formula for the general term \(a_{n}\) of the sequence, assuming that the pattern of the first few terms continues. $$\left\\{1,-\frac{1}{3}, \frac{1}{9},-\frac{1}{27}, \frac{1}{81}, \ldots\right\\}$$

Step-by-Step Solution

Answer

The general term of the sequence is \(a_n = (-1)^{n+1}\left(\frac{1}{3}\right)^{n-1}\).

Step 1: Identify the Pattern

Observe the sequence \[ 1, -\frac{1}{3}, \frac{1}{9}, -\frac{1}{27}, \frac{1}{81}, \ldots \]The terms alternate in sign and the absolute value of each term follows a geometric sequence with a common ratio of 1/3.

Step 2: Establish the General Term

Since the sequence alternates in sign, this can be represented by \((-1)^n\). The absolute values of the terms follow a geometric sequence with the first term 1 and common ratio \(\frac{1}{3}\). Thus, the nth term of the positive sequence is given by\( \left( \frac{1}{3} \right)^{n-1} \).

Step 3: Combine Elements into Formula

Combine the elements from Steps 1 and 2 to form the formula for the general term, remembering to start indexing from \(n=1\): \[ a_n = (-1)^{n+1} \left( \frac{1}{3} \right)^{n-1} \].

Step 4: Simplify Formula

Simplify the expression from Step 3 to ensure it matches the pattern observed in the sequence: \[ a_n = (-1)^{n+1} \left( \frac{1}{3} \right)^{n-1}= (-1)^{n+1}\left(3^{1-n}\right) \].

Key Concepts

Geometric Sequence

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. These sequences are very predictable, and they can be identified with each term having a consistent multiplication factor. For instance, in the exercise sequence:

• First term is 1
• Second term is \(-\frac{1}{3}\), obtained by multiplying 1 (the previous term) by \(-\frac{1}{3}\)
• Third term is \(\frac{1}{9}\), and so on.

Although the signs alternate, the absolute values of the terms arise from multiplying by \(\frac{1}{3}\), demonstrating a geometric progression.

Alternating Sign

An alternating sign sequence is a sequence where the signs of the terms alternate between positive and negative. This pattern can be easily described mathematically by the expression \((-1)^n\).When \(n\) is odd, \((-1)^n\) equals -1, producing a negative term. When \(n\) is even, \((-1)^n\) equals 1, resulting in a positive term.In our sequence, the alternating sign is crucial for forming the correct sequence and contributes to the general term formula structure:\[ a_n = (-1)^{n+1} \left( \frac{1}{3} \right)^{n-1} \].This ensures the terms switch sign each time \(n\) increments.

Common Ratio

The common ratio in a geometric sequence is the constant factor you multiply each term by to get to the next term. In the discussed exercise, the common ratio is identified as \(\frac{1}{3}\).Finding the common ratio involves dividing one term by the previous term. For example:

• \(-\frac{1}{3} / 1 = -\frac{1}{3}\)
• \(\frac{1}{9} / -\frac{1}{3} = -\frac{1}{3}\)

This shows that each consequent term is always \(\frac{1}{3}\) of the previous term in absolute value, sustaining the geometric sequence. The alternating sign means that while the magnitudes form a geometric sequence, the signs change regularly. The common ratio is essential for determining the structure of the sequence's general term, allowing us to maintain the same ratio throughout the sequence's progression.