Problem 22 - Algebra and Trigonometry

Question

Use the Principle of Mathematical Induction to show that the given statement is true for all natural numbers \(n\). $$ n(n+1)(n+2) \text { is divisible by } 6 $$

Step-by-Step Solution

Answer

By mathematical induction, \(n(n+1)(n+2)\) is divisible by 6 for all natural numbers \(n\).

Step 1: Base Case

Verify the statement for the initial natural number, which is usually 1. Evaluate the expression when \(n = 1\). Calculate \(1(1+1)(1+2)\) and determine if it is divisible by 6.

Step 2: Base Case Calculation

For \(n = 1\):\[ 1(1+1)(1+2) = 1 \cdot 2 \cdot 3 = 6 \] Since 6 is divisible by 6, the base case holds true.

Step 3: Inductive Hypothesis

Assume that the statement holds true for some arbitrary natural number \(k\), meaning \(k(k+1)(k+2)\) is divisible by 6. This is our inductive hypothesis.

Step 4: Inductive Step

Prove that if the statement holds for \(n = k\), then it also holds for \(n = k+1\). Evaluate \((k+1)((k+1)+1)((k+1)+2)\) and show it is also divisible by 6.

Step 5: Expansion

For \(n = k+1\):\[ (k+1)((k+1)+1)((k+1)+2) = (k+1)(k+2)(k+3) \]

Step 6: Factorization

Rewrite \((k+1)(k+2)(k+3)\) and argue its divisibility by 6. Note that this product contains three consecutive integers. One of these integers must be divisible by 2, and one must be divisible by 3, because every three consecutive integers include at least one multiple of each of these primes.

Step 7: Conclusion of Induction

Since \((k+1)(k+2)(k+3)\) is divisible by both 2 and 3, it is divisible by 6. Hence, if the statement is true for \(n = k\), it is true for \(n = k+1\). By mathematical induction, the given statement is true for all natural numbers \(n\).

Key Concepts

Divisibility

In mathematics, divisibility refers to whether one number can be divided by another without leaving a remainder. For example, 10 is divisible by 2 because 10 divided by 2 equals 5, which is an integer. When we talk about the divisibility of expressions, we mean whether the result of an expression is divisible by a given number. In this exercise, we need to show that for any natural number \( n \), the expression \( n(n+1)(n+2) \) is divisible by 6. To do this, we need to prove that for any \( n \), the product of three consecutive numbers is always divisible by both 2 and 3, and thus by 6.

Base Case

The base case in mathematical induction is the first step where you verify that the given statement works for the initial value. Usually, this initial value is 1. In our exercise, the base case checks whether \( 1(1+1)(1+2) \) is divisible by 6. Let's calculate it:
For \( n = 1 \), the expression becomes \( 1 \times 2 \times 3 = 6 \).
Since 6 is divisible by 6, our base case holds true. This step is crucial as it sets the foundation for the inductive step, ensuring the statement is true for the first natural number.

Inductive Step

The inductive step is where we assume that the statement holds true for some arbitrary natural number \( k \), and then prove it for \( k+1 \). This assumption that the statement is true for \( k \) is known as the inductive hypothesis. In our exercise, we assume that \( k(k+1)(k+2) \) is divisible by 6. Then, we need to prove that \( (k+1)(k+2)(k+3) \) is also divisible by 6.
Notice that \( (k+1)(k+2)(k+3) \) are three consecutive numbers, just like \( k(k+1)(k+2) \). One of these three consecutive numbers is always divisible by 2, and another is always divisible by 3. Therefore, their product is divisible by 6.

Natural Numbers

Natural numbers are the set of positive integers starting from 1, which include numbers like 1, 2, 3, and so on. They do not include negative numbers, fractions, or decimals. These numbers are key in many mathematical principles, including induction.
When using mathematical induction, the principles are applied to natural numbers to prove statements for all possible cases in this infinite set. For this exercise, we show that our statement holds true for all natural numbers starting from 1.