Chapter 12.4, Problem 22AYU

To determine

To prove: The given statement $n\left(n + 1\right)\left(n + 2\right)$ is divisible by 6 is true for all natural numbers $n$ using the Principle of Mathematical Induction.

Answer to Problem 22AYU

As the statement is true for the natural number $\left(k + 1\right)$ terms, hence the statement is true for all natural numbers.

Explanation of Solution

Given:

Statements says the series $n\left(n + 1\right)\left(n + 2\right)$ is divisible by 6 is true for all natural number.

Formula used:

The Principle of Mathematical Induction

Suppose that the following two conditions are satisfied with regard to a statement about natural numbers:

CONDITION I: The statement is true for the natural number 1.

CONDITION II: If the statement is true for some natural number $k$, it is also true for the next natural number $k + 1$. Then the statement is true for all natural numbers.

Proof:

Consider the statement

$n\left(n + 1\right)\left(n + 2\right)$ is divisible by 6    -----(1)

Step 1: Show that statement (1) is true for the natural number $n = 1$.

That is $1 \left(1 + 1\right)\left(1 + 2\right) = 6$ is divisible by 6. Hence the statement is true for the natural number $n = 1$.

Step 2: Assume that the statement is true for some natural number $k$.

That is $k\left(k + 1\right)\left(k + 2\right)$ is divisible by 6   -----(2)

Step 3: Prove that the statement is true for the next natural number $k + 1$.

That is, to prove that $\left(k + 1\right)\left(k + 2\right)\left(k + 3\right)$ is divisible by 2

Consider $\left(k + 1\right)\left(k + 2\right)\left(k + 3\right) = k\left(k + 1\right)\left(k + 2\right) + 3 \left(k + 1\right)\left(k + 2\right)$

From equation (2), first term in the above equation is divisible by 6 and second term is a multiple of three and product of two consecutive integers.

It is a known fact that the product of two consecutive integers is even. So the consecutive integers $\left(k + 1\right)\left(k + 2\right)$ is even that is it’s a multiple of 2.

The product of multiple of 3 and multiple of 2 is the multiple of 6. Therefore the second term is divisible by 6.

Hence $\left(k + 1\right)\left(k + 2\right)\left(k + 3\right)$ is divisible by 6.

As the statement is true for the natural number $\left(k + 1\right)$ terms, hence the statement is true for all natural numbers.