Problem 14 - College Algebra

Question

In Exercises 11-24, use mathematical induction to prove that each statement is true for every positive integer \(n.\) $$3+6+9+\dots+3 n=\frac{3 n(n+1)}{2}$$

Step-by-Step Solution

Answer

By using mathematical induction, we prove that the formula \(3+6+9+\dots+3n=\frac{3n*(n+1)}{2}\) holds for every positive integer \(n\). The base case was verified and the induction step was proven by assuming the formula is true for \(n=k\)

Step 1: Base Case

Let's first show that the formula holds when \(n=1\). According to the formula, \(3*1*(1+1)/2=3\). This shows that when we put \(n=1\), both sides are equal, and thus the formula is correct for \(n=1\).

Step 2: Assumption Step

Next, suppose or assume that the formula is true for some \(n=k\), i.e., it's assumed that \(3+6+9+\dots+3 k=\frac{3k*(k+1)}{2}\).

Step 3: Inductive Step

Now, let's prove it for \(n=k+1\). We will add the next term \(3*(k+1)\) to both sides of our assumption step and simplify. So it would be \(3+6+9+\dots+3k+3*(k+1)=\frac{3k*(k+1)}{2} + 3*(k+1)\). We'll simplify this to find the right hand side of the equation being equal to \(\frac{3*(k+1)*(k+2)}{2}\). Thus, if the formula is true for \(n=k\), it is also true for \(n=k+1\).

Step 4: Conclusion

Since the formula holds for \(n=1\) (i.e., the base case) and \(n=k+1\) (i.e., the inductive step), we can conclude by the principle of mathematical induction that the given formula is valid for every positive integer \(n\). For the detailed derivation, refer to the above steps.

Key Concepts

Base Case

In mathematical induction, the base case is the starting point of the proof. We use it to verify that our statement or formula works for the smallest possible value of the variable, typically when the variable is equal to 1. In the context of proving a formula involving positive integers with mathematical induction, we start by checking the base case. In the given exercise, the base case is when \( n = 1 \).

• We plug \( n = 1 \) into both sides of the formula.
• Perform the calculations to ensure they equal each other.
• If both sides match, the base case is verified as true.

In our example, we have \( 3 = \frac{3 \times 1 \times (1 + 1)}{2} \), which simplifies to 3. Thus, the formula holds true for \( n = 1 \). Successfully proving the base case is a crucial first step in any inductive proof.

Inductive Step

The inductive step is the part of mathematical induction that involves proving a statement holds for \( n = k + 1 \), given that it holds for \( n = k \). This forms the heart of the mathematical induction process.
To perform an inductive step:

• Assume the formula is true for some arbitrary positive integer \( k \). This is known as the inductive hypothesis.
• Add the next element in the sequence to both sides of the assumption equation.
• Simplify and verify that the formula holds for \( n = k + 1 \).

In our example, we assume \( 3 + 6 + 9 + \dots + 3k = \frac{3k(k + 1)}{2} \) is true. We then prove it for \( n = k + 1 \) by adding \( 3(k + 1) \) to both sides and simplifying them into the form \( \frac{3(k + 1)(k + 2)}{2} \). If both sides are valid, the inductive step is successfully completed.

Positive Integer

In the world of mathematics, a positive integer is a whole number greater than zero. These numbers have significant relevance in mathematical problems and proofs because they are simple yet foundational.
Positive integers:

• Include numbers such as 1, 2, 3, and so on.
• Do not have fractions, decimals, or negative signs.
• Are common in sequences, series, and induction problems.

In the exercise we are discussing, positive integers like \( n \), \( k \), and \( k + 1 \) play a crucial role. They are integral to the sequences and sums for which we are trying to prove validity using mathematical induction, and they provide the structure for equations like \( 3n(n+1)/2 \). Understanding the concept of positive integers is key to grasping the basics of such proofs.

Sum of Series

A sum of series refers to the total obtained by adding each number in a sequence. In mathematical induction, showing that the sum of a series matches a given formula is a common task.
To calculate a sum of a series:

• Recognize the pattern or rule that generates the series.
• Apply the rule to each term in the sequence to verify the formula.
• Use techniques, such as induction, to prove the formula holds for all relevant terms.

In our specific exercise, the series consists of terms like 3, 6, 9, up to 3n. The sum of these terms is captured by the formula \( \frac{3n(n+1)}{2} \). We demonstrate it is true for every positive integer \( n \) using mathematical induction. Understanding how to manipulate and prove sums of series is crucial in higher mathematics.