Question

For any positive integer n,prove that n^3 -n is divided by 6.​

Answer 1

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Answer 2

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$For \: any \: positive \: integer \: n, \: prove \: that \\ n^3 -n \: is \: divided \: by \: 6.$

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Let us consider:-

$a = {n}^{3} - n$

$a = n( {n}^{2} - 1)$

$a = n(n + 1)(n - 1)$

Assumptions:-

1. Out of three (n – 1), n, (n + 1) one must be even, so a is divisible by 2.

2. (n – 1) , n, (n + 1) are consecutive integers thus as proved a must be divisible by 3.

From (1) and (2) a must be divisible by 2 × 3 = 6

Thus, n³ – n is divisible by 6 for any positive integer n.

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