Problem 20 - College Algebra

Question

In Exercises 11-24, use mathematical induction to prove that each statement is true for every positive integer \(n.\) $$\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\dots+\frac{1}{2^{n}}=1-\frac{1}{2^{n}}$$

Step-by-Step Solution

Answer

The given formula \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\dots+\frac{1}{2^{n}}=1-\frac{1}{2^{n}}\) is successfully proven to be true for all positive integers \(n\) using mathematical induction.

Step 1: Base Case: \(n=1\)

We insert \(n=1\) into the given formula: \(\frac{1}{2^{1}}=1-\frac{1}{2^{1}}\), which simplifies to \(\frac{1}{2}=\frac{1}{2}\). This is true, so our base case is confirmed.

Step 2: Induction Hypothesis

Assume the given formula is true for \(n=k\). This gives us the equality: \(\frac{1}{2^{1}}+\frac{1}{2^{2}}+\dots+\frac{1}{2^{k}}=1-\frac{1}{2^{k}}\)

Step 3: Inductive Step: Prove for \(n=k+1\)

We want to prove the formula is true for \(n=k+1\). So we'll consider the left side of the equation for \(n=k+1\): \(\frac{1}{2^{1}}+\frac{1}{2^{2}}+\dots+\frac{1}{2^{k}} + \frac{1}{2^{k+1}}\). Replace the sum up to \(n=k\) by \(1 - \frac{1}{2^{k}}\) (using Induction Hypothesis): \(1 - \frac{1}{2^{k}} + \frac{1}{2^{k+1}}\). This simplifies to: \(1 - \frac{1}{2^{k+1}}\), which is exactly the right-hand side for \(n=k+1\). So, the formula holds for \(n=k+1\) if it's true for \(n=k\).

Step 4: Inductive Conclusion

By Principle of Mathematical Induction, the formula \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\dots+\frac{1}{2^{n}}=1-\frac{1}{2^{n}}\) is true for all positive integers \(n\).

Key Concepts

Base Case

The base case in mathematical induction serves as the starting point of our proof. It's a bit like the anchor for the mathematical induction process. For the given problem, the base case begins with checking if our formula holds true for the smallest positive integer, usually, this is where we set \(n=1\).
Let's delve into the formula:

• Given formula: \(\frac{1}{2} = 1 - \frac{1}{2}\)
• Plug in \(n=1\)
• This simplifies to \(\frac{1}{2} = \frac{1}{2}\)

The true statement confirms that our formula holds for \(n=1\). Establishing this step is crucial because it sets the precedent for all other cases.
Without a true base case, the whole induction could potentially fall apart.

Inductive Hypothesis

In the next step of mathematical induction, we introduce the inductive hypothesis. This step is crucial because it represents the "assume it works" step.
Here’s how we use this concept:

• Assume the formula is true for a given integer \(n = k\):
• \(\frac{1}{2^{1}} + \frac{1}{2^{2}} + \cdots + \frac{1}{2^{k}} = 1 - \frac{1}{2^{k}}\)

The idea here is similar to balancing on a tightrope. We assume the formula holds at this point, providing a stepping stone to prove it for the next integer. The assumption is our bridge to continue the proof forwards.

Inductive Step

The inductive step is often seen as the most critical piece of the mathematical induction process. This is where the magic happens, allowing us to leap from one case to the next.
To perform the inductive step, follow these steps:

• Assume the formula is true for \(n=k\).
• Now, prove it is true for \(n = k + 1\):
• Start with the left-hand side for \(n=k+1\): \ \(\frac{1}{2^{1}} + \frac{1}{2^{2}} + \cdots + \frac{1}{2^{k+1}}\)
• Substitute the known formula for \(n=k\) with \(1 - \frac{1}{2^{k}}\)
• Add the next term \(\frac{1}{2^{k+1}}\): \ \(1 - \frac{1}{2^{k}} + \frac{1}{2^{k+1}}\)

After simplifying, you should reach the form: \ \(1 - \frac{1}{2^{k+1}}\)
If this aligns with the formula for \(n=k+1\), our proof is good to go. The magic of this step is showing any case being true leads to the next being true.

Recursive Proof

The recursive proof is another way to view or visualize the induction process. While the term "recursive" often implies a function calling itself, in the context of this proof, it's a way to think about repeatedly applying the pattern or logic.
In essence, mathematical induction itself is recursive; we’re proving a base, then showing how each case leads directly to the next, ad infinitum.

• Start from our foundation, the base case.
• Then with the inductive hypothesis, assume all steps are true up to a certain point.
• With the inductive step, we confirm the pattern continues indefinitely.

This recursive nature is like dominoes falling — if one domino falls (or a step holds true), it causes the next to fall (hold true) as well, forever pushing the boundary forward. The recursive proof process ensures the infinite extent of proofs, leveraging the established pattern.