For any positive integer n, prove that n^3 – n is divisible by 6.

+1 vote
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asked Nov 26, 2017 in Mathematics by Golu (37,045 points) 19 169 604

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

1 Answer

+2 votes
answered Nov 26, 2017 by anukriti (13,536 points) 5 10 42
selected Nov 26, 2017 by sarthaks
 
Best answer

Solution:
n3 – n = n (n2 – 1) = n (n – 1) (n + 1)
Therefore, n3 – n is product of three consecutive positive integers, where n is any positive integer.
Since one out of every two consecutive integers is divisible by 2.

Therefore, The product n3 – n is divisible by 2.
Since one out of every three consecutive integers is divisible by 3.
Therefore, The product n3 – n is divisible by 3.
Any number which is divisible by 2 and 3 is also divisible by 6.
Hence, The product n3 – n is divisible by 6.

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