Problem 27 - Algebra and Trigonometry

Question

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

Using mathematical induction, it has been proven that 6 is a factor of \(n(n+1)(n+2)\) for every positive integer \(n\). This was done by establishing a base case and showing that the property of being divisible by 6 holds for \(k+1\) if it holds for an arbitrary positive integer \(k\).

Step 1: Base Case

Check if 6 is a factor of \(1(1+1)(1+2)\). Simplifying this results in 6, which indeed is divisible by 6.

Step 2: Induction Hypothesis

Assume the statement holds true for some positive integer \(k\). This means that 6 is a factor of \(k(k+1)(k+2)\).

Step 3: Induction Step

Prove that if the hypothesis holds for \(k\), it also holds for \(k+1\). A key step in doing this is to expand \((k+1)((k+1)+1)((k+1)+2)\), which simplifies to \((k+1)(k+2)(k+3)\). Note that among \(k+1\), \(k+2\), and \(k+3\), at least one of these three numbers is divisible by 2 and at least one is divisible by 3. Hence, their product is divisible by 6. This completes the induction and the statement is proven.

Key Concepts

Factor of a Number

A factor of a number is a smaller number that can divide the given number without leaving any remainder. For instance, the number 6 is actually composed of the factors 1, 2, 3, and 6.
To understand factors better, think about how 6 can be multiplied by other numbers to reach a total:

• 1 times 6 equals 6
• 2 times 3 equals 6

When something is a factor, like 6 in this mathematical proof, it indicates that our final answer can be shared evenly by 6.
In the example problem, we need to prove that 6 divides evenly into the expression \(n(n+1)(n+2)\) for any positive integer \(n\). By showing this consistency, we establish that 6 is indeed a factor across all possible cases.

Positive Integers

Positive integers are simply all whole numbers greater than zero. These include numbers like 1, 2, 3, 4, and so on, extending infinitely. They do not include zero or any negative numbers.
In the context of induction proofs, positive integers are important because we often need to establish that something is true for all of these numbers. The starting point, or base case, is usually the smallest positive integer, which is 1.
By proving something is true for this number, and then showing that it holds for any number \(k\), we can inductively prove it applies to all positive integers. This method is part of why mathematical induction is a powerful tool in mathematics.

Proof Techniques

Proof techniques are essential tools in mathematics to establish the validity of statements. Among these techniques, mathematical induction is particularly powerful when working with statements that need to be proven true for all positive integers.
When using mathematical induction, we typically follow these steps:

• Start with the base case: Prove that the statement is true for the initial value, usually 1.
• Then, assume it's true for an arbitrary positive integer \(k\), which is our induction hypothesis.
• Next, using this hypothesis, prove that the truth of the statement extends to \(k+1\).
• If these steps are successfully completed, the statement is shown to hold for all positive integers.

This method hinges on the idea that if you can step from one integer to the next, maintaining truth, then the truth must hold for all integers along that path. It's akin to climbing an infinite ladder where each step supports you securely.