Problem 27 - College Algebra

Question

In Exercises 25-34, use mathematical induction to prove that each statement is true for every positive integer \(n.\) 6 is a factor of \(n(n+1)(n+2)\).

Step-by-Step Solution

Answer

Yes, by using mathematical induction, it is proven that 6 is a factor of \(n(n+1)(n+2)\) for every positive integer \(n\).

Step 1: Base Case (n = 1)

The base case tests whether the statement is true for \(n = 1\). Substituting \(n = 1\) we get \(1(1+1)(1+2) = 6\), which clearly has 6 as a factor.

Step 2: Inductive Hypothesis (n = k)

Assume the statement is true for some integer k, that means 6 is a divider for \(k(k+1)(k+2)\).

Step 3: Inductive Step (n = k + 1)

To show the statement is true for \(k + 1\), you need to find out whether 6 is a factor of \((k+1)((k+1)+1)((k+1)+2) = (k+1)(k+2)(k+3)\). Considering the inductive hypothesis and factorizing \((k+1)(k+2)(k+3)\) as \((k+1)(k+2)k + (k+1)(k+2)\cdot 3\), it is seen that both terms on the right side are divisible by 6. So, it's true that 6 is also a factor \((k+1)(k+2)(k+3)\). Therefore, by the principle of mathematical induction, 6 is a factor of \(n(n+1)(n+2)\) for all positive integer \(n\).

Key Concepts

Base Case

Mathematical induction begins with the base case, which is the foundation of the entire proof.
In our exercise, the base case is when the positive integer \( n \) equals 1. This step verifies if the statement holds true for this initial value.
More specifically, we substitute \( n = 1 \) into the expression \( n(n+1)(n+2) \).

• When \( n = 1 \), the expression becomes \( 1(1+1)(1+2) = 6 \).
• Evidently, 6 is a factor of itself, confirming that the base case holds true.

Proving the base case is crucial because it serves as a stepping stone for further proof.
If the base case isn't valid, the rest of the induction process might break down, leading to incorrect conclusions.

Inductive Hypothesis

After confirming the base case, the next step involves setting up the inductive hypothesis.
This involves assuming that the statement is true for a certain, but arbitrary, positive integer \( k \).
In our example, the hypothesis states that 6 is a factor of \( k(k+1)(k+2) \).
This means we've presumed the truth of the statement for this specific \( k \).

• The inductive hypothesis doesn't prove anything by itself.
• Its role is to aid in proving the statement for the next integer, \( k+1 \).

The assumption here sets the stage for the crucial next phase: extending the truth from \( k \) to \( k+1 \).
This hypothetical assumption is a standard and necessary part of the mathematical induction process.

Inductive Step

The inductive step is where the magic of mathematical induction truly unfolds.
This is the phase where we show that if the statement holds true for \( n = k \), then it must also hold true for \( n = k + 1 \).
In our exercise, this involves demonstrating that 6 is a factor of \((k+1)(k+2)(k+3)\).

• Start by expressing \((k+1)(k+2)(k+3)\).
• Factorize it as \((k+1)(k+2)k + 3(k+1)(k+2)\).
• Using the inductive hypothesis, since \( k(k+1)(k+2) \) is divisible by 6 by assumption, this implies similar divisibility for \((k+1)(k+2)(k+3)\).

This step is crucial because it confirms the extendability of the truth from the integer \( k \) to \( k+1 \).
Thus, when both the base case and the inductive step are successfully verified, the principle of mathematical induction ensures that the statement is true for all positive integers \( n \).