Problem 12 - A Graphical Approach to Precalculus with Limits

Question

Use mathematical induction to prove each statement. Assume that \(n\) is a positive integer. $$5 \cdot 6+5 \cdot 6^{2}+5 \cdot 6^{3}+\cdots+5 \cdot 6^{n}=6\left(6^{n}-1\right)$$

Step-by-Step Solution

Answer

The statement is proven true using mathematical induction.

Step 1: Understanding the Induction Principle

Mathematical induction involves two main steps: the base case and the inductive step. We first verify that the given statement holds true for the initial value (usually, when \( n = 1 \)). Then, we assume the statement is true for \( n = k \) and prove it for \( n = k+1 \).

Step 2: Base Case Verification

Let's check if the equation holds for \( n = 1 \). We substitute into the left and right sides:Left side: \( 5 \cdot 6 = 30 \)Right side: \( 6(6^1 - 1) = 6(6 - 1) = 30 \)Both sides equal, thus the base case is verified for \( n = 1 \).

Step 3: Inductive Hypothesis Assumption

Assume that for some positive integer \( k \), the statement holds true:\[ 5 \cdot 6 + 5 \cdot 6^2 + 5 \cdot 6^3 + \cdots + 5 \cdot 6^k = 6(6^k - 1) \]

Step 4: Inductive Step Proof

We need to prove the statement for \( n = k+1 \):\[ 5 \cdot 6 + 5 \cdot 6^2 + 5 \cdot 6^3 + \cdots + 5 \cdot 6^k + 5 \cdot 6^{k+1} = 6(6^{k+1} - 1) \]Starting with the left side, add \( 5 \cdot 6^{k+1} \) to the inductive hypothesis:\[ \text{LHS} = 6(6^k - 1) + 5 \cdot 6^{k+1} \]Distribute and simplify:\[ = 6 \cdot 6^k - 6 + 5 \cdot 6^{k+1} \]\[ = 6^{k+1} - 6 + 5 \cdot 6^{k+1} \]Combine terms:\[ = 6^{k+1} + 5 \cdot 6^{k+1} - 6 \]\[ = 6^{k+1}(1 + 5) - 6 \]\[ = 6 \cdot 6^{k+1} - 6 \]\[ = 6(6^{k+1} - 1) \]Since the left-hand side equals the right-hand side, the statement is true for \( n = k+1 \).

Step 5: Conclusion

By the principle of mathematical induction, since the base case is true and the inductive step holds, the statement \( 5 \cdot 6 + 5 \cdot 6^2 + 5 \cdot 6^3 + \cdots + 5 \cdot 6^n = 6(6^n - 1) \) is true for all positive integers \( n \).

Key Concepts

Base Case Verification

The foundation of mathematical induction lies in verifying the base case. This is the first and critical step in induction, showing the statement is true for the initial value, typically when \( n = 1 \). This step provides a "starting point" that the entire proof builds upon.

In the exercise, our statement is
\( 5 \cdot 6 + 5 \cdot 6^2 + 5 \cdot 6^3 + \cdots + 5 \cdot 6^n = 6(6^n - 1) \).
For the base case, we substitute \( n = 1 \).

On the left side, we compute:
\( 5 \cdot 6 = 30 \).

On the right side, we calculate:
\( 6(6^1 - 1) = 6(6 - 1) = 30 \).

Since both sides equal 30, the base case verification is successful. This confirms that the statement holds true when \( n = 1 \), ensuring a solid foundation for the induction process.

Inductive Hypothesis

In the induction process, the inductive hypothesis is a critical assumption that the statement is true for some arbitrary positive integer \( k \). This does not mean that it's definitely true at this stage, but rather, we presume its truth for the sake of the proof.

For our specific exercise, the inductive hypothesis is:
\[ 5 \cdot 6 + 5 \cdot 6^2 + 5 \cdot 6^3 + \cdots + 5 \cdot 6^k = 6(6^k - 1) \].

This assumption underpins the next part of the induction process, where we aim to prove the statement for \( k+1 \). The concept is akin to laying a stepping stone that paves the way to prove the statement holds true for the subsequent integer, thus establishing a chain reaction in the proof.

Inductive Step Proof

The inductive step proof is the heart of mathematical induction. It involves proving that if the statement holds for \( n = k \), then it must also hold for \( n = k+1 \).

To accomplish this, we begin with our inductive hypothesis:
\[ 5 \cdot 6 + 5 \cdot 6^2 + 5 \cdot 6^3 + \cdots + 5 \cdot 6^k = 6(6^k - 1) \].

Next, we add the next term, \( 5 \cdot 6^{k+1} \), to both sides:

• Left Side (LHS): Start with the inductive hypothesis and add the new term:
  \[ LHS = 6(6^k - 1) + 5 \cdot 6^{k+1} \].


• Combine like terms and factor:
  \[ = 6^{k+1} - 6 + 5 \cdot 6^{k+1} \]
  \[ = 6^{k+1}(1 + 5) - 6 \]
  \[ = 6 \cdot 6^{k+1} - 6 \]
  \[ = 6(6^{k+1} - 1) \].

As both sides of the equation match, our inductive step proof is complete. This confirms that if the statement is true for \( n = k \), it's also true for \( n = k+1 \). Therefore, by the principle of induction, the original proposition holds true for all positive integers.