n^3 – n is divisible by 6, for each natural number n ≥ 2.

Question

Prove the statement the Principle of Mathematical Induction :

n³ – n is divisible by 6, for each natural number n ≥ 2.

Answer 1

Let P(n): n³ – n is divisible by 6, for each natural number n> 2.

Now, P(2): (2)³ - 2 = 6, which is divisible by 6.

Hence, P(2) is true.

Let us assume that, P(n) is true for some natural number n = k.

P(k): k³ – k is divisible by 6

or k³ - k = 6m,  m ∈ N .......(i)

Now, we have to prove that P(k + 1) is true.

P(k+ 1) : (k+ l)³ - (k+ 1)

= k³ + 1 + 3k(k + l) - (k+ 1)

= k³ + 1 +3k² + 3k - k - 1 

= (k³ - k) + 3k(k + 1)

= 6m + 3 k(k +1)         (using (i))

Above is divisible by 6. (∴ k(k + 1) is even)

Hence, P(k + 1) is true whenever P(k) is true.

So, by the principle of mathematical induction P(n) is true for any natural number n,n≥ 2.