Question

for any positive integer n, prove that n3(cube) - n is divisible by 6.

Answer 1

Sol:
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 principle of  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.