Problem 9 - Algebra for College Students

Question

List all terms of each finite sequence. \(b_{n}=\frac{(-1)^{n}}{n}\) for \(1 \leq n \leq 10\)

Step-by-Step Solution

Answer

-1, 1/2, -1/3, 1/4, -1/5, 1/6, -1/7, 1/8, -1/9, 1/10

Step 1 - Understanding the sequence formula

The given sequence is defined by the formula: \( b_{n} = \frac{(-1)^n}{n} \). This tells us how to generate each term based on the value of \( n \).

Step 2 - Identify the range for n

We need to find terms for \( n \) in the range from 1 to 10 inclusive. This means \( n \) will take on the values 1, 2, 3, ..., 10.

Step 3 - Calculate each term

We now substitute each value of \( n \) into the formula: For \( n = 1 \): \( b_{1} = \frac{(-1)^{1}}{1} = -1 \) For \( n = 2 \): \( b_{2} = \frac{(-1)^{2}}{2} = \frac{1}{2} \) For \( n = 3 \): \( b_{3} = \frac{(-1)^{3}}{3} = -\frac{1}{3} \) For \( n = 4 \): \( b_{4} = \frac{(-1)^{4}}{4} = \frac{1}{4} \) For \( n = 5 \): \( b_{5} = \frac{(-1)^{5}}{5} = -\frac{1}{5} \) For \( n = 6 \): \( b_{6} = \frac{(-1)^{6}}{6} = \frac{1}{6} \) For \( n = 7 \): \( b_{7} = \frac{(-1)^{7}}{7} = -\frac{1}{7} \) For \( n = 8 \): \( b_{8} = \frac{(-1)^{8}}{8} = \frac{1}{8} \) For \( n = 9 \): \( b_{9} = \frac{(-1)^{9}}{9} = -\frac{1}{9} \) For \( n = 10 \): \( b_{10} = \frac{(-1)^{10}}{10} = \frac{1}{10} \)

Key Concepts

Sequence Formula

Understanding the sequence formula is the first step in solving this exercise. The given sequence formula is \( b_{n} = \frac{(-1)^n}{n} \). This formula helps you generate each term in the sequence based on the value of \(n\). To break it down:

• \((-1)^n\) is a term that alternates between +1 and -1 depending on whether \(n\) is even or odd.
• \(\frac{1}{n}\) is a term that takes the reciprocal of the integer \(n\).

Combining these, for a given \(n\), you either get a positive or negative fraction. For example, when \(n = 1\), the fraction is negative because \((-1)^1 = -1\), and when \(n = 2\), the fraction is positive because \((-1)^2 = 1\). This switching between positive and negative values continues for all terms in the sequence.

Range of n

Next, let's talk about the range of \(n\). The problem specifies that \(n\) ranges from 1 to 10 inclusive, denoted by \(1 \leq n \leq 10\). This means that we're looking for values of \(n\) including all integers between and including 1 and 10. This step-by-step method helps us identify the exact terms we need to calculate. So for this exercise, \(n\) takes on the values:

• 1,
• 2,
• 3,
• 4,
• 5,
• 6,
• 7,
• 8,
• 9,
• 10

Each of these values will be used in the sequence formula to calculate the corresponding sequence terms.

Term Calculation

Now that we know the range of \(n\), the next step is calculating the terms. We substitute each value of \(n\) into the sequence formula:\( b_{n} = \frac{(-1)^n}{n} \). Let's do it step-by-step:

• For \(n = 1\): \( b_{1} = \frac{(-1)^{1}}{1} = -1 \)
• For \(n = 2\): \( b_{2} = \frac{(-1)^{2}}{2} = \frac{1}{2} \)
• For \(n = 3\): \( b_{3} = \frac{(-1)^{3}}{3} = -\frac{1}{3} \)
• For \(n = 4\): \( b_{4} = \frac{(-1)^{4}}{4} = \frac{1}{4} \)
• For \(n = 5\): \( b_{5} = \frac{(-1)^{5}}{5} = -\frac{1}{5} \)
• For \(n = 6\): \( b_{6} = \frac{(-1)^{6}}{6} = \frac{1}{6} \)
• For \(n = 7\): \( b_{7} = \frac{(-1)^{7}}{7} = -\frac{1}{7} \)
• For \(n = 8\): \( b_{8} = \frac{(-1)^{8}}{8} = \frac{1}{8} \)
• For \(n = 9\): \( b_{9} = \frac{(-1)^{9}}{9} = -\frac{1}{9} \)
• For \(n = 10\): \( b_{10} = \frac{(-1)^{10}}{10} = \frac{1}{10} \)

Following this method ensures we calculate every term correctly.

Negative and Positive Terms

Why do the sequence terms alternate between negative and positive? This happens because of the \((-1)^n\) part of the formula. Let's go deeper:

• \(((-1)\) raised to an odd power (e.g., \(n = 1, 3, 5\)) gives \(-1\).
• \(((-1)\) raised to an even power (e.g., \(n = 2, 4, 6\)) gives \(1\).

This alternating pattern of signs is critical. It helps to understand why the sequence cannot just be fractions, but must include signs as well. Testing the formula with different values of \(n\) can help you see this pattern:

• For \(n = 5\), because 5 is odd, the term is negative \(-\frac{1}{5}\).
• For \(n = 6\), because 6 is even, the term is positive \(\frac{1}{6}\).

Observing these alternations helps you predict and verify if computations are correct as you progress.