Question

For any positive integer n, prove that (n^3– n) is divisible by 6...please answer q.no.7

Answer 1

Answer:

This can be done by two methods

METHOD - 1

We have

n³ - n = n(n² - 1) = n(n+1)(n-1)

(n-1) , n , (n+1) are consecutive positive integers. Product of three consecutive positive integers is divisible by 6.

So, n(n-1)(n+1) is divisible by 6

=> n³ - n is divisible by 6

METHOD - 2

Every positive integer is of the form 6q + r, 0 <= r < b

i.e. r = 0,1,2,3,4,5

i.e. every positive integer is either one of the forms 6q, 6q+1, ... , 6q+5

Now, put n = 6q, 6q + 1, ... , 6q + 5. For all those values of n, you will find that n³ - n is divisible by 6

This proves the statement.

You can use any of the above methods to solve this question.