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Chapter 2: Problem 30

Prove by induction that $$ 1+r+r^{2}+\cdots+r^{k}+\cdots+r^{n}=\frac{1-r^{n+1}}{1-r} $$

Short Answer

Expert verified
Proving this formula involves three main steps: establishing a valid base case, assuming that the property holds for a specific \(k\) (the induction hypothesis), and verifying that if it holds for \(k\), it must also hold for \(k+1\) (the induction step). Following these steps carefully demonstrates that the formula is indeed valid.

Step by step solution

01

Base Case

In mathematical induction, we start by proving the base case. Here the base case would be for \( n = 0 \), which means we only have to verify that \(1 = \frac{1 - r^{0+1}}{1 - r}\). If you simplify that, it indeed equals 1, since \( 1- r \) in the denominator cancels out with \(1 - r\) in the numerator.
02

Induction Hypothesis

The induction hypothesis is where we assume that the property holds true for a certain \(k\). So we assume that \(1 + r + r^2 + ... + r^k = \frac{1 - r^{k+1}}{1 - r}\) is true.
03

Induction Step

The final step of mathematical induction is to prove that if the property holds for \(k\), then it must also hold for \(k+1\). Therefore, let's think about what happens when we add \(r^{k+1}\) to both sides of our equation. We get \((1 + r + r^2 + ... + r^k) + r^{k+1} = \frac{1 - r^{k+1}}{1 - r} + r^{k+1}\). This is the same as \((1 + r + r^2 + ... + r^{k} + r^{k+1}) = \frac{1 - r^{k+1} + r^{k+1}(1 - r)}{1 - r}\). If you simplify the right side, you can see that it indeed equals \( \frac{1 - r^{k+2}}{1 - r}\). This is the result we wanted, so we can conclude that the induction step holds, and the formula is proved.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Base Case
The base case is the starting point in a proof by mathematical induction. It is essential to verify it carefully, as the subsequent steps depend on it being true. In our example, we consider the base case when \( n = 0 \). This means we only examine the initial term of the series. The equation simplifies to verifying if:
  • \( 1 = \frac{1 - r^{0+1}}{1 - r} \)
The expression on the right side becomes \( \frac{1 - r}{1 - r} \), which simplifies further to \( 1 \), confirming our base case is true. Successfully validating the base case means we are ready to proceed to the next phase of induction.
Induction Hypothesis
In this step, we assume the formula holds for an arbitrary case, typically noted as \( k \). This assumption is known as the induction hypothesis. For our problem, we assume:
  • \( 1 + r + r^2 + \cdots + r^k = \frac{1 - r^{k+1}}{1-r} \)
This assumption serves as a temporary truth that aids us in demonstrating that the property is valid for the next term \((k+1)\). It's important to clearly state this hypothesis as all following work hinges on this temporary assumption being true for the step that follows.
Induction Step
The induction step is where the crux of the proof happens. During this step, using the induction hypothesis, we show that the formula also holds for \( k+1 \). Essentially, we prove:
  • If the formula is true for \( k \), it is also true for \( k+1 \)
  • We consider \( (1 + r + r^2 + \cdots + r^k) + r^{k+1} = \frac{1 - r^{k+1}}{1 - r} + r^{k+1} \)
The task is to simplify this expression and see if it conforms to the pattern. By combining terms and factoring, we recognize that the expression simplifies to:
  • \( \frac{1 - r^{k+1} + r^{k+1}(1 - r)}{1-r} \)
This results in:
  • \( \frac{1 - r^{k+2}}{1 - r} \)
Achieving this form confirms that the property holds for \( k+1 \). Thus, the process of mathematical induction ensures the formula is true for all natural numbers \( n \).

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