Problem 30 - College Algebra

Question

In Exercises 25-34, use mathematical induction to prove that each statement is true for every positive integer \(n.\) $$\sum_{i=1}^{n} 7 \cdot 8^{i}=8\left(8^{n}-1\right)$$

Step-by-Step Solution

Answer

After proving the base case for \(n=1\), and assuming the formula works for \(n=k\), we showed that the formula also works for \(n=k+1\). Therefore, by mathematical induction, the formula holds for all positive integers \(n\).

Step 1: Base Case

The first step in mathematical induction is to prove the base case. We show that the formula holds for \(n=1\). So, we substitute \(n=1\) into both sides of the equation. \(\sum_{i=1}^{1} 7 \cdot 8^{i}=7 \cdot 8^{1}=7 \cdot 8=56\) and \(8(8^{1}-1)=8(8-1)=8 \cdot 7=56\). Both left and right sides are equals, so base case is satisfied.

Step 2: Inductive Step - Assumption

Next, we assume that the formula holds for \(n=k\), this is also known as the inductive hypothesis. We thus have \(\sum_{i=1}^{k} 7 \cdot 8^{i}=8(8^{k}-1)\) for some positive integer \(k\).

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

The final step is to prove that the formula holds for \(n=k+1\) given that it holds for \(n=k\). We calculate the left hand side (LHS) for \(n=k+1\): \(\sum_{i=1}^{k+1} 7 \cdot 8^{i}=\sum_{i=1}^{k} 7 \cdot 8^{i} + 7*8^{k+1}\). But according to the inductive hypothesis \(\sum_{i=1}^{k} 7 \cdot 8^{i} = 8(8^{k}-1)\), substituting this in the LHS we get \(8(8^{k}-1) + 7*8^{k+1}\). Now calculate the right hand side (RHS) for \(n=k+1\): \(8(8^{k+1}-1)\). We need to show that the LHS equals the RHS. Let's do some algebra \[8(8^{k}-1) + 7*8^{k+1}=8*8^{k} - 8 + 7*8^{k} * 8 = 15*8^{k} - 8 = 8(8^{k+1}-1)\] The LHS does indeed equals the RHS. Therefore, our formula works for \(n=k+1\) when it works for \(n=k\). We've completed the proof by induction.

Key Concepts

Base Case

The base case is the starting point of any mathematical induction problem. You begin by validating the statement for the smallest positive integer, often denoted as \(n=1\). In our example, the formula \(\sum_{i=1}^{n} 7 \cdot 8^{i}=8\left(8^{n}-1\right)\) is checked for \(n=1\).

Here, substituting \(n=1\) into both sides:

• The left-hand side (LHS) becomes \(7 \cdot 8^1 = 56\).
• The right-hand side (RHS) becomes \(8(8^1 - 1) = 8 \cdot 7 = 56\).

Both LHS and RHS equate to 56, confirming that the base case holds true.

This initial step establishes a foundation upon which the inductive hypothesis and step will build. It's crucial because if the smallest case doesn’t hold true, neither will any subsequent cases.

Inductive Hypothesis

The inductive hypothesis involves assuming that the proposition holds for a specific integer \(n=k\). This assumption allows you to set the stage for proving the next step. It is often phrased as: if the statement is true for \(n=k\), we must show it is also true for \(n=k+1\).

So, assuming the formula \(\sum_{i=1}^{k} 7 \cdot 8^{i}=8(8^{k}-1)\) holds for some integer \(k\), we use this as our inductive hypothesis. This assumption is not about proving it at the moment but using it as a crutch to reach the next step: proving \(n=k+1\).

Keep in mind that this assumption is vital to the inductive process. It’s like climbing a ladder: you assume the previous rung can support your weight as you reach for the next rung.

Inductive Step

The inductive step is where the magic of mathematical induction happens. This step involves taking the assumption from the inductive hypothesis and proving that if it's true for \(n=k\), then it is true for \(n=k+1\).

For our formula, the goal is to transition from \(n=k\) to \(n=k+1\). Here's how this is done:

• Start with the hypothesis: \(\sum_{i=1}^{k} 7 \cdot 8^{i}=8(8^{k}-1)\).
• Add the next term in the series to both sides: \(\sum_{i=1}^{k+1} 7 \cdot 8^{i} = \sum_{i=1}^{k} 7 \cdot 8^{i} + 7 \cdot 8^{k+1}\).
• Substitute the hypothesis into this equation: \(8(8^{k}-1) + 7 \cdot 8^{k+1}\).
• Show that it simplifies to the RHS for \(n=k+1\): \(8(8^{k+1}-1)\).

This algebraic validation ensures our formula holds for \( k+1 \) assuming it holds for \( k \), thereby climbing another rung on our mathematical ladder. This step demonstrates the beautiful logical framework of induction.

Proof by Induction

Proof by induction is a powerful and widely-used mathematical technique. It is similar to knocking over a row of dominoes, where pushing the first one causes the entire sequence to fall.

This process consists of three central components:

• Base Case: Prove that a statement is true for the initial, often smallest, value of \(n\).
• Inductive Hypothesis: Assume the statement is true for an arbitrary positive integer \(n=k\).
• Inductive Step: Demonstrate that if it holds for \(n=k\), it holds for \(n=k+1\).

Returning to the domino analogy, the base case is equivalent to knocking over the first domino. The inductive hypothesis assumes that each falling domino in turn will cause the next to fall, and the inductive step is the reassurance that this will indeed happen.

In conclusion, once the base case and inductive step are verified, you have a watertight proof that the statement holds for all positive integers \(n\). Proof by induction is a crucial tool in mathematics, providing a way to prove propositions extending indefinitely.