For any positive integer n prove that n³-n is divisible by 6?

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For any positive integer n prove that n³-n is divisible by 6?

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For any positive integer n prove that n³-n is divisible by 6?

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.

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