Prove that n^3-n is divisible by 6

Given

n3 – n is divisible by 6 ∀ n ∈ N

Proof

We will prove the given statement by induction

STEP 1

n = 1

n3 – n = 13 – 1 = 0

0 is divisible by 6. Therefore, the statement is true for n = 1

STEP 2

Let the given statement be true for n = k

k3 – k = 6x

Now, we need to prove that if the statement is true for n = k then it is also true for n = k + 1

(k + 1)3 – (k + 1) = k3 + 1 + 3k2 + 3k – k – 1

= k3 – k + 3k2 + 3k

= 6x + 3(k2 + k)

The term k2 + k is divisible by 2, why?

If k = EVEN, then k2 = EVEN. We know, EVEN + EVEN = EVEN

If k = ODD, then k2 = ODD, We know, ODD + ODD = EVEN

(k + 1)3 – (k + 1) is divisible by 6 if k3 – k is divisible by 6.

Therefore, we can say that if the given statement is true for n = k, then it is also true for n = k + 1. Hence, by the principle of mathematical induction, the given statement is true ∀ n ∈ N.

Leave a Comment

Your email address will not be published. Required fields are marked *