Problem 29 - College Algebra

Question

In Exercises 25-34, use mathematical induction to prove that each statement is true for every positive integer \(n.\) $$\sum_{i=1}^{n} 5 \cdot 6^{i}=6\left(6^{n}-1\right)$$

Step-by-Step Solution

Answer

The statement \(\sum_{i=1}^{n} 5 \cdot 6^{i}=6\left(6^{n}-1\right)\) holds true for all positive integers \(n\), as proven by the method of mathematical induction.

Step 1: Base Step

We begin by showing the proposition holds for the base case, \(n=1\). So, we replace \(n\) with 1 in the statement: \(5 \cdot 6^{1} = 6(6^{1} - 1)\), which simplifies to: \(30 = 30\). So, the proposition holds true for \(n=1\).

Step 2: Inductive Step - Inductive Hypothesis

Now we assume the proposition holds for \(n=k\). This is called the Inductive Hypothesis. So we assume: \(\sum_{i=1}^{k} 5 \cdot 6^{i}=6\left(6^{k}-1\right)\).

Step 3: Inductive Step - Verification

Next, we prove the proposition holds for \(n=k+1\) given it holds for \(n=k\). We start by adding \(5 \cdot 6^{k+1}\) to both sides of the inductive hypothesis: \( \sum_{i=1}^{k+1} 5 \cdot 6^{i}=6\left(6^{k}-1\right) + 5 \cdot 6^{k+1}\). We need to prove that right-hand side equals to \(6(6^{k+1} - 1)\). If we simplify the right-hand side, it becomes: \(6 \cdot 6^{k+1} - 6 + 5 \cdot 6^{k+1}\), which becomes: \(6(6^{k+1}-1)\), which is what we were supposed to prove.

Step 4: Conclusion

Since the proposition is true for \(n=1\) and assuming the proposition is true for \(n=k\) allows us to prove it's also true for \(n=k+1\), by the Principle of Mathematical Induction, the proposition is true for all positive integers \(n\).

Key Concepts

Series Summation

Understanding series summation is crucial when dealing with sequences and mathematical induction. A series is essentially the sum of the terms of a sequence. In this problem, we have a series represented by

• \( \sum_{i=1}^{n} 5 \cdot 6^{i} \)

which means we add up the terms from \(i=1\) to \(i=n\). Each term in this series follows a pattern, specifically \(5 \cdot 6^{i}\).

Series summation can simplify complex sequences into manageable expressions. The goal here is to prove that this series equals \(6(6^{n} - 1)\) for every positive integer \(n\). This task involves using mathematical induction and understanding each term’s contribution to the series as \(i\) increases.

Inductive Hypothesis

The inductive hypothesis is a key component in mathematical induction, referring to the assumption step of the proof.

After proving the base case (the simplest instance), we assume the proposition is true for some arbitrary positive integer \(k\). In our problem, this means assuming:

• \( \sum_{i=1}^{k} 5 \cdot 6^{i} = 6(6^{k} - 1) \)

This assumption helps us "leap" to the next case, \(n=k+1\), to prove it still holds true. Think of it as a stepping stone: if the stone at \(k\) is solid, we can step confidently to \(k+1\). This assumes confidence in "if it works for \(k\), it'll work for the next step." Remember, it's not a guess—it's a strategic step assuming consistency across iterations.

Base Case

The base case is the beginning of every mathematical induction process. It verifies the proposition for the initial value, ensuring it holds true from the start.

In this example, we examine the base case \(n=1\). By substituting \(n\) with 1, we get:

• \(5 \cdot 6^{1} = 6(6^{1} - 1)\)
• which simplifies to \(30 = 30\).

This confirms that the given proposition is valid for \(n=1\). Without this foundational validity, moving on to other positive integers would be baseless.
The base case essentially jumps us into the inductive process, setting the initial anchor point for further exploration.

Algebraic Proof

Algebraic proof is about demonstrating the equality or validity of equations using algebraic manipulations. In our problem, once we have the assumption from the inductive hypothesis, we need to verify it for \(n=k+1\).

We begin by modifying our initial assumption to include one more term:

• \( \sum_{i=1}^{k+1} 5 \cdot 6^{i} = 6(6^{k} - 1) + 5 \cdot 6^{k+1} \)

Our task is to algebraically simplify this expression to match \(6(6^{k+1} - 1)\), proving the same structural form as earlier.

By careful rearrangement and simplification through distributive properties, the expression transforms satisfactorily into:

• \(6 \cdot 6^{k+1} - 6 + 5 \cdot 6^{k+1} = 6(6^{k+1} - 1)\)

Thus, through algebraic methods, we validate the correctness for \(n=k+1\), thereby proving our statement for all positive integers.