Problem 14 - A Graphical Approach to Precalculus with Limits

Question

Use mathematical induction to prove each 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 proven by mathematical induction for all positive integers \(n\).

Step 1: Base Case

To start with mathematical induction, verify the base case. Here, the smallest positive integer in our context is \(n = 1\). For \(n = 1\), the left-hand side of the equation is:\[7 \cdot 8 = 56\]The right-hand side is:\[8 \times (8^1 - 1) = 8 \times (8 - 1) = 8 \times 7 = 56\]Since both sides are equal, the base case holds true.

Step 2: Inductive Hypothesis

Now, assume that the statement holds true for some positive integer \(k\). That is, assume:\[7 \cdot 8 + 7 \cdot 8^2 + 7 \cdot 8^3 + \dots + 7 \cdot 8^k = 8(8^k - 1)\]This assumption is called the inductive hypothesis.

Step 3: Inductive Step

We need to prove the statement for \(n = k + 1\). Begin from the left-hand side of the equation:\[7 \cdot 8 + 7 \cdot 8^2 + 7 \cdot 8^3 + \dots + 7 \cdot 8^k + 7 \cdot 8^{k+1}\]Using the inductive hypothesis, replace the sum up to \(k\):\[= 8(8^k - 1) + 7 \cdot 8^{k+1}\]

Step 4: Simplify the Expression

Simplify the right-hand side of the equation:\[= 8 imes 8^k - 8 + 7 \cdot 8^{k+1}\]Factor out \(8\) and combine like terms:\[= 8^1 \cdot 8^k + 7 \cdot 8^{k+1} - 8 \]\[= 8^{k+1} + 56 \cdot 8^k - 8 \]

Step 5: Rewrite and Factor the Result

Notice that \(8^{k+1}\) can be factored out:\[= 8^{k+1} - 8 = 8(8^{k+1} - 1)\]This matches the right-hand side of the equation for \(n = k+1\):\[7 \cdot 8 + 7 \cdot 8^2 + \cdots + 7 \cdot 8^{k+1} = 8(8^{k+1} - 1)\]

Step 6: Conclusion

Since the base case holds, and the assumption that the step from \(k\) to \(k + 1\) has been proven, by the principle of mathematical induction, the statement is true for all positive integers \(n\).

Key Concepts

Base Case

In mathematical induction, the base case is the initial step where we verify that the statement holds true for the smallest integer value, usually denoted as 1. This step serves as the foundation for the induction process. In our problem, the given equation is checked for \(n = 1\).

On the left side, we calculate \(7 \cdot 8 = 56\). Similarly, on the right, we compute \(8(8^1 - 1) = 8(8 - 1) = 8 \cdot 7 = 56\). Both results are equal, confirming that the base case is correct. Without this confirmation, the entire induction process cannot proceed.

In mathematical induction, establishing the base case is crucial. It proves that the statement is true for the first number, and sets the stage for proving that it holds for all subsequent numbers.

Inductive Hypothesis

The inductive hypothesis forms the bridge between the base case and the generalization process. Here, you assume that the statement is true for a specific integer \(k\). This means we take it on faith that our statement works for this value.

In our specific example, we assume:

• \(7 \cdot 8 + 7 \cdot 8^2 + 7 \cdot 8^3 + \dots + 7 \cdot 8^k = 8(8^k - 1)\)

It is important to remember that this assumption is only for the sake of convenience. This hypothesis does not prove anything on its own, but it’s essential for the next step. The goal is to use this hypothesis to demonstrate that the statement is also valid for \(n = k + 1\).

Inductive Step

The inductive step is where the magic of induction truly happens. In this step, you prove that if the statement holds true for \(n = k\) (as assumed in the inductive hypothesis), then it must also hold true for \(n = k + 1\).

Starting with:

• \(7 \cdot 8 + 7 \cdot 8^2 + \dots + 7 \cdot 8^k + 7 \cdot 8^{k+1} = 8(8^{k+1} - 1)\)

Use the inductive hypothesis to replace the initial portion of the sum:

• \(= 8(8^k - 1) + 7 \cdot 8^{k+1}\)

Simplify and combine terms:

• \(= 8^{k+1} - 8 + 7 \cdot 8^{k+1}\)
• \(= 8^{k+1} + 56 \cdot 8^k - 8\)

Factor and simplify further to confirm that it matches the right-hand side of the equation:

• \(= 8(8^{k+1} - 1)\)

This shows that if the statement is true for \(n = k\), it is also true for \(n = k + 1\).

Proof by Induction

Proof by induction is a powerful mathematical tool used to verify that a statement or formula is correct for every integer, starting from a base case. This method works through two main steps:

• Base Case: The foundation where we show the statement is true for the beginning point.
• Inductive Step: The logical leap showing that if it's true for one integer, it follows for the next.

This technique effectively acts as a domino effect. Once you've proven the first case and shown that each case follows the previous, every case must indeed be true.

In our given exercise, confirming the base case at \(n = 1\) and the inductive step from \(n = k\) to \(n = k + 1\) proves the statement for all integers greater than or equal to 1. Thus, using proof by induction, we successfully demonstrate that the proposition is valid for every positive integer.