Problem 8 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 1-26: Prove that the statement is true for every positive integer \(n\). $$ (-1)^{1}+(-1)^{2}+(-1)^{3}+\cdots+(-1)^{n}=\frac{(-1)^{n}-1}{2} $$

Step-by-Step Solution

Answer

The statement is proven by mathematical induction.

Step 1: Understand the Problem Statement

The problem asks us to prove that the sum of the series \((-1)^1 + (-1)^2 + \cdots + (-1)^n\) is equal to \(\frac{(-1)^n - 1}{2}\) for every positive integer \(n\). This requires us to show why this equation holds true for all positive integers.

Step 2: Check Base Case for Mathematical Induction

The base case starts with \(n = 1\). The left-hand side (LHS) becomes \((-1)^1 = -1\) and the right-hand side (RHS) becomes \(\frac{(-1)^1 - 1}{2} = \frac{-1 - 1}{2} = -1\). Thus, the base case holds as \(-1 = -1\).

Step 3: Set Up Inductive Hypothesis

Assume the statement is true for some positive integer \(k\). That is, \((-1)^1 + (-1)^2 + \cdots + (-1)^k = \frac{(-1)^k - 1}{2}\). This is our induction hypothesis, which we need to prove for \(n = k + 1\).

Step 4: Prove the Inductive Step

We need to show that if the statement holds for \(n = k\), then it must also hold for \(n = k + 1\). Consider the left-hand side for \(n = k + 1\): \((-1)^1 + (-1)^2 + \cdots + (-1)^k + (-1)^{k+1}\). Using the inductive hypothesis, this becomes \(\frac{(-1)^k - 1}{2} + (-1)^{k+1}\).

Step 5: Simplify and Verify

Substitute and simplify: 1. Substitute: \(= \frac{(-1)^k - 1}{2} + (-1)^{k+1}\)2. Combine: \(= \frac{(-1)^k - 1 + 2(-1)^{k+1}}{2}\)3. Rewrite: \(= \frac{(-1)^{k+1} - 1}{2}\).Since this equals the right-hand side \(\frac{(-1)^{k+1} - 1}{2}\), the inductive step is verified.

Step 6: Conclusion

By the principle of mathematical induction, since both the base case and the inductive step are true, the statement is proved for every positive integer \(n\). Therefore, the equation \((-1)^1 + (-1)^2 + \cdots + (-1)^n = \frac{(-1)^n - 1}{2}\) is true for all positive integer \(n\).

Key Concepts

Series Summation

Series summation involves adding up a sequence of numbers. In our problem, we look at the series \((-1)^1 + (-1)^2 + \cdots + (-1)^n\). This type of series alternates signs between positive and negative.
This is because for any integer \(n\), \((-1)^n\) gives either \(1\) or \(-1\) depending on whether \(n\) is even or odd.

• An even index, like \(n=2\), results in \((-1)^2 = 1\).
• An odd index, like \(n=3\), results in \((-1)^3 = -1\).

This pattern creates a repeating sequence of 1 and -1, impacting the summation. As \(n\) increases, it becomes crucial to determine if \(n\) is even or odd to simplify the series effectively.

Base Case

The base case is the starting point of mathematical induction. Essentially, it tests the smallest value in a logical sequence, typically \(n=1\).
For our series, the base case is when \(n = 1\):

• Left-hand side (LHS): \((-1)^1 = -1\).
• Right-hand side (RHS): \(\frac{(-1)^1 - 1}{2} = \frac{-1 - 1}{2} = -1\).

Here, the LHS equals the RHS, confirming that the statement holds true for \(n=1\).
This crucial step in induction demonstrates that our formula is valid at its initial position, setting the stage for proving subsequent steps.

Inductive Hypothesis

The inductive hypothesis is an assumption that the given statement is true for a specific value, usually denoted as \(n = k\).
In our exercise, we assume that for some positive integer \(k\), the statement is valid:

• The series: \((-1)^1 + (-1)^2 + \cdots + (-1)^k = \frac{(-1)^k - 1}{2}\)

This assumption helps us move forward to prove the next incremental step in our sequence. The "inductive hypothesis" is a vital bridge that connects our proof of the base case to demonstrating that the pattern holds in subsequent steps.

Inductive Step

The inductive step aims to verify that if the statement holds for \(n = k\), then it must also hold for \(n = k + 1\).
This step involves utilizing the inductive hypothesis and proving its truth for \(k+1\).

• For \(n = k + 1\), the series becomes: \((-1)^1 + (-1)^2 + \cdots + (-1)^k + (-1)^{k+1}\).
• Apply the hypothesis: \(\frac{(-1)^k - 1}{2} + (-1)^{k+1}\).
• Simplify: \(\frac{(-1)^k - 1 + 2(-1)^{k+1}}{2} = \frac{(-1)^{k+1} - 1}{2}\).

Since we derive \(\frac{(-1)^{k+1} - 1}{2}\), equivalent to the RHS for \(n = k + 1\), our inductive step is verified.
Therefore, this step ensures continuity of the pattern established in the base case and inductive hypothesis, proving the statement for all positive integers \(n\).