Problem 10 - College Algebra with Modeling and Visualization

Question

Use mathematical induction to prove the statement. Assume that \(n\) is a positive integer. $$ 7 \cdot 8+7 \cdot 8^{2}+7 \cdot 8^{3}+\dots+7 \cdot 8^{n}=8\left(8^{n}-1\right) $$

Step-by-Step Solution

Answer

The statement is true for all positive integers using mathematical induction.

Step 1: Base Case

Check if the statement is true for the smallest positive integer, which is typically 1 in induction proof. Substitute \( n = 1 \) into the equation.Left side: \( 7 \cdot 8 = 56 \)Right side: \( 8(8^1 - 1) = 8(8 - 1) = 8 \cdot 7 = 56 \)Both sides are equal, thus the base case holds true.

Step 2: Inductive Hypothesis

Assume that the statement holds true for some integer \( n = k \), that is:\[ 7 \cdot 8 + 7 \cdot 8^2 + ... + 7 \cdot 8^k = 8(8^k - 1). \]

Step 3: Inductive Step

We need to show that the statement is also true for \( n = k + 1 \). Start with the left side of the equation:\[ 7 \cdot 8 + 7 \cdot 8^2 + \dots + 7 \cdot 8^k + 7 \cdot 8^{k+1}. \]Using the inductive hypothesis:\[ 8(8^k - 1) + 7 \cdot 8^{k+1}. \]Simplifying that expression:\( 8 \cdot 8^k - 8 + 7 \cdot 8^{k+1} = 8^{k+1} - 8 + 56 \times 8^k. \)\( = 8 \cdot (8^{k+1} - 1). \)Simplified correctly, this shows that the induction step holds.

Step 4: Conclusion

Since the base case holds and the inductive step has been proven, by mathematical induction, the statement is true for all positive integers \( n \).

Key Concepts

Base Case

In the world of mathematical induction, the base case acts as your starting point, setting the foundation for your proof. When you want to prove that a statement holds for all positive integers, you must first check if it works for the simplest case, usually when the integer is 1. For our given problem, we substitute 1 for \( n \) in both sides of the statement. The left side becomes \( 7 \cdot 8 = 56 \) and the right side simplifies to \( 8\times(8^1 - 1) = 8 \times 7 = 56 \). These calculations reveal that both sides are indeed equal. This successful match confirms the base case and assures us that the statement holds true for \( n = 1 \). Hence, the proof stands strong at the very first step, ready to move forward.

Inductive Hypothesis

Next up is the inductive hypothesis, a pivotal part of the induction process. Imagine it as assuming your statement works for a certain integer \( n = k \), hoping this assumption will help prove it for \( n = k + 1 \). Here, we assume:

• \( 7 \cdot 8 + 7 \cdot 8^2 + \ldots + 7 \cdot 8^k = 8(8^k - 1) \)

Think of it as a leap of faith, trusting that the statement is true for any general \( k \). This assumption sets the stage for testing the next step. However, remember, this assumption alone doesn’t prove anything yet; it's more of a bridge to help cross over to the next part of the proof. The hypothesis is necessary because it allows us to use known information to extrapolate to the next number in sequence.

Inductive Step

The inductive step is where the magic happens. This is where we show that if the statement is true for \( n = k \), then it must also be true for \( n = k + 1 \). Start with the left side of the statement under this new condition:

• \( 7 \cdot 8 + 7 \cdot 8^2 + \ldots + 7 \cdot 8^k + 7 \cdot 8^{k+1} \).

Using our inductive hypothesis, we know the sum up to \( 7\cdot 8^k \) is \( 8(8^k - 1) \). Adding \( 7 \cdot 8^{k+1} \) gives:

• \( 8(8^k - 1) + 7 \cdot 8^{k+1} \).

Simplify this expression to show that it equals \( 8(8^{k+1} - 1) \), validating that adding one more term continues the pattern. This critical step confirms that if the statement holds for \( n = k \), it also holds for \( n = k + 1 \). Successfully completing this step ensures that the statement is true for all positive integers, validating the entire induction process.