Question

for any positive integer n, prove that n3 - n is divisible by 6

Answer 1

By the mathematical induction we have to prove this.

⇒  Given that S(n) = n3-n divisible by 6.

Let n =1 then we get '0'

which is divisible by 6.

  ∴ S(1) is true.

Let us assume that n = k

 S(k) = k3- k

which is divisible by 6.

∴ S(k) is true.

∴ (k3-k) / 6 = m ( integer )

  (k3-k)  = 6m

   k3= 6m +k --------→(1)

now we have to prove that n = k+1

⇒ (k+1)3 - (k+1)

⇒ (k3+3k2+3k+1) - (k+1)

subsitute equation (1) in above equation then

⇒ 6m +k+3k2+2k

⇒ 6m +3k2+k

⇒ 6m +3k(k+1)  ( ∴k(k+1) = 2p is an even number p is natural number)

⇒ 6m +3x2p

⇒ 6(m +p)

∴which is divisible by 6

s(k+1) is true.

By the mathematical induction it is true for n∈N.

Answer 2

Answer:

n3 – n = n(n2 – 1) = n(n+1)(n – 1) = (n – 1)n(n+1) = product of three consecutive positive integers.

Now, we have to show that the product of three consecutive positive integers is divisible by 6.

We know that any positive integer n is of the form 3q, 3q + 1 or 3q + 2 for some positive integer q.

Now three consecutive positive integers are n, n + 1, n + 2.

Case I. If n = 3q.

n(n + 1) (n + 2) = 3q(3q + 1) (3q + 2)

But we know that the product of two consecutive integers is an even integer.

∴ (3q + 1) (3q + 2) is an even integer, say 2r.

⟹ n(n + 1) (n + 2) = 3q × 2r = 6qr, which is divisible by 6.

Case II. If n = 3n + 1.

∴ n(n + 1) (n + 2) = (3q + 1) (3q + 2) (3q + 3)

= (even number say 2r) (3) (q + 1)

= 6r (q + 1),

which is divisible by 6.

Case III. If n = 3q + 2.

∴ n(n + 1) (n + 2) = (3q + 2) (3q + 3) (3q + 4)

= multiple of 6 for every q

= 6r (say),

which is divisible by 6.

Hence, the product of three consecutive integers is divisible by 6.