Question

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

Answer 1

Answer :-

let us consider

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

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

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

Assputions :-

• out of the three n -1 , n , n +1 one must be even so a is divisible by 2 .
• n -1 , n , n +1 are consecutive integer thus as proved a must be divisible by 3 .
• from (1) and (2) a must be divisible by 2 x 3 = 6 thus , n ^3-n is divisible by 6 for any positive integer "n " .

Thanks

Answer 2

Answer:

So to make it divisible by 6 , the no. should be divisible by both 2&3.

= n3-n= 12

= 3n= 12+n

= 3n-n= 12

= 2n= 12

= n= 12/2

6

proving----

3n= 6×3

18

n= 6

3n-n= 2n

2n= 12(2×6)

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