Problem 14 - A Graphical Approach to College Algebra

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 true by mathematical induction for all positive integers \( n \).

Step 1: Base Case

Firstly, we need to test the base case where \( n = 1 \). Substitute \( n = 1 \) into both sides of the equation.The left-hand side (LHS) becomes: \[ 7 \cdot 8^1 = 56 \]The right-hand side (RHS) becomes:\[ 8 \cdot (8^1 - 1) = 8 \cdot (8 - 1) = 8 \cdot 7 = 56 \]Since LHS equals RHS, the base case holds true.

Step 2: Inductive Hypothesis

Assume the statement is true for \( n = k \), i.e.,\[ 7 \cdot 8 + 7 \cdot 8^2 + 7 \cdot 8^3 + \, ... \, + 7 \cdot 8^k = 8(8^k - 1) \]This is our inductive hypothesis.

Step 3: Inductive Step

We need to show that if the statement is true for \( n = k \), it must also be true for \( n = k+1 \).Start with the original series and include the next term:\[ 7 \cdot 8 + 7 \cdot 8^2 + 7 \cdot 8^3 + \, ... \, + 7 \cdot 8^k + 7 \cdot 8^{k+1} \]By the inductive hypothesis, the first part is equal to:\[ 8(8^k - 1) \]Now, add the term \( 7 \cdot 8^{k+1} \):\[ 8(8^k - 1) + 7 \cdot 8^{k+1} \]

Step 4: Simplify and Prove for \( n = k+1 \)

Factor 8 out of the expression:\[ 8(8^k - 1) + 7 \cdot 8^{k+1} = 8 \, \times \, 8^k - 8 \, + \, 8 \, \times \, 7 \, \times \, 8^k \]Combine like terms:\[ 8^{k+1} + 56 \, \times \, 8^k - 8 = 8^{k+1} + 8 \, \times \, 7 \, \times \, 8^k - 8 \]Simplify to:\[ 8(8^{k+1} - 1) \]Thus, the statement is true for \( n = k+1 \). This completes the inductive step.

Step 5: Conclusion

Since we have proved the base case and the inductive step, by the principle of mathematical induction, the statement holds true for all positive integers \( n \).

Key Concepts

Understanding Algebra in Induction Proofs

Algebra is a core component of mathematics that deals with variables, symbols, and the rules for manipulating these. When used in proofs, especially in mathematical induction, algebra allows us to rewrite equations, combine like terms, and simplify expressions to reveal deeper relationships between numbers and operations. In the context of mathematical induction, algebra is particularly useful in:

• Simplifying expressions: This can involve factoring, expanding, or rearranging terms to uncover patterns that are not immediately obvious. For example, in the exercise we simplify the expression by factoring 8 out of several terms.
• Assigning and using variables: In an induction proof, we often use a variable such as \( n \) to represent some integer position in a sequence or pattern. We then use algebraic manipulation to show how this variable affects the entire equation or set of conditions.
• Working through iterative steps: As expressions become increasingly complex with higher values of \( n \), algebra ensures accuracy and clarity, breaking down complicated expressions into simpler, actionable parts.

Algebra in induction enables us to bridge the assumption of truth for \( n = k \) and prove it for \( n = k+1 \), which is essential for completing an induction proof.

Geometric Series and Its Role in the Proof

A geometric series is a series of terms where each term after the first is found by multiplying the previous one by a fixed, non-zero number. This fixed number is known as the 'common ratio.' In the given exercise, the geometric series involves powers of 8: \( 8^1, 8^2, 8^3, \ldots, 8^n \). Each term can be represented in the form \( 7 \cdot 8^k \), establishing a pattern of growth that is key to solving this problem by induction. Here are a few reasons why understanding geometric series is vital in this proof:

• Recognizing patterns: Observing that each term follows a consistent multiplicative pattern helps establish the formula we wish to prove with induction.
• Simplifying calculations: The sum of a geometric series can often be simplified using known formulas. For example, the sum of the series is written in a way that allows for factoring and subsequent simplification.
• Forming the inductive hypothesis: The induction process relies on understanding how the sum up to \( n \) terms behaves to predict the addition of the \( n+1 \) term.

In the context of the exercise, recognizing the series as geometric allows the use of these properties to create a compact representation of the sum, ultimately simplifying the entire proof process.

Proof Techniques: Mathematical Induction

Mathematical induction is a proof technique that is particularly potent for proving statements about all natural numbers. It's like a domino effect, where you show that if one domino (number) falls, they all will. For the task of proving a statement with induction, you follow these essential steps:

• Base Case: Verify that the statement is true for the first number in the series, usually \( n = 1 \). This establishes a starting point or initial condition for the induction.
• Inductive Hypothesis: Assume the statement is true for some arbitrary positive integer \( k \). This assumption lets you build on a proven case.
• Inductive Step: Demonstrate that if the statement holds for \( n = k \), it must also hold for \( n = k+1 \). This links all successive terms together.

In our example, we use these steps to confirm that \( 7 \cdot 8 + 7 \cdot 8^2 + 7 \cdot 8^3 + \ldots + 7 \cdot 8^n = 8(8^n -1) \) holds for every positive integer \( n \). Thus, induction not only showcases the truth of a given mathematical sequence or property but also the logical flow underpinning that truth. It's a systematic way to prove statements that involve progression from one step to the next.