Question

prove that n cube - n is divisible by 6 for every positive integer

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

Step-by-step explanation:

n³ - n = n (n² - 1) = n (n - 1) (n + 1)

Whenever a number is divided by 3, the remainder obtained is either 0 or 1 or 2.

∴ n = 3p or 3p + 1 or 3p + 2, where p is some integer.

If n = 3p, then n is divisible by 3.

If n = 3p + 1, then n – 1 = 3p + 1 –1 = 3p is divisible by 3.

If n = 3p + 2, then n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) is divisible by 3.

So, we can say that one of the numbers among n, n – 1 and n + 1 is always divisible by 3.

⇒ n (n – 1) (n + 1) is divisible by 3.

Similarly, whenever a number is divided 2, the remainder obtained is 0 or 1.

∴ n = 2q or 2q + 1, where q is some integer.

If n = 2q, then n is divisible by 2.

If n = 2q + 1, then n – 1 = 2q + 1 – 1 = 2q is divisible by 2 and n + 1 = 2q + 1 + 1 = 2q + 2 = 2 (q + 1) is divisible by 2.

So, we can say that one of the numbers among n, n – 1 and n + 1 is always divisible by 2.

⇒ n (n – 1) (n + 1) is divisible by 2.

Since, n (n – 1) (n + 1) is divisible by 2 and 3.

∴ n (n-1) (n+1) = n³ - n is divisible by 6.( If a number is divisible by both 2 and 3 , then it is divisible by 6)

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