Question

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Q. For any positive integer n, prove that n3 – n is divisible by 6.​

Answer 1

$\huge\bold\red\bigstar$QUESTION:-

For any positive integer n, prove that n3 – n is divisible by 6.

$\huge\bold\red\bigstar$ANSWER:-

$\implies$ a = n³-n.

$\implies$ n(n²-1)

$\implies$ n(n-1) (n+1)

$\implies$ (n-1) n(n+1)

$\implies$ 1)now out of the three (n-1) ,n and (n+1) one must be even do a divisible by 6.

$\implies$ 2) also (n - 1),n and (n + 1) are three consecutive integers thus as proved a must be divisible by 3.

$\implies$ From (1) and (2)

$\implies$ a must be divisible by 2 × 3 = 6

$\huge\bold\red\bigstar$Hence n³ - n is divisible by 6 for any positive integer n.

Answer 2

Step-by-step explanation:

⟹ a = n³-n.

⟹ n(n²-1)

⟹ n(n-1) (n+1)

⟹ (n-1) n(n+1)

⟹ 1)now out of the three (n-1) ,n and (n+1) one must be even do a divisible by 6.

⟹ 2) also (n - 1),n and (n + 1) are three consecutive integers thus as proved a must be divisible by 3.

⟹ From (1) and (2)

⟹ a must be divisible by 2 × 3 = 6