Question

for any positive integer n, prove that n3-n is divisible by 6.

Answer 1

$<b>$let's suppose that


a = n³ - n


= n(n² - 1)


= n(n - 1)(n + 1)


= (n - 1)n (n + 1)


1) Now out of three (n - 1),n and (n + 1) one must be even so a is divisible by 2


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


Hence n³ - n is divisible by 6 for any positive integer n.


Thanks

Have a colossal day ahead

Be Brainly

Answer 2

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

n^3 – n = n(n^2 –1)

= n(n + 1)(n – 1)

= (n – 1) n(n + 1)

= Product of three consecutive positive integers  Now, we have to show that the product of  three consecutive positive integers is divisible  by 6.

We know that any positive integer a is of the  form 3q, 3q + 1 or 3q + 2 for some integer q.

Let a, a + 1, a + 2 be any three consecutive  integers.

If a = 3q

a(a + 1)(a + 2) = 3q(3q + 1)(3q + 2)

= 3q (2r)

= 6qr, which is divisible by 6.

If a = 3q + 1

a(a + 1)(a + 2) = (3q + 1)(3q + 2)(3q + 3)

= (2r) (3)(q + 1)

= 6r(q + 1) which is divisible by 6

If a = 3q + 2

a(a + 1)(a + 2) = (3q + 2)(3q + 3)(3q + 4)  = multiple of 6 for every  q = 6r (say) which is divisible by 6.

Therefore the product of three consecutive  integers is divisible by 6.