Question

Hi..
Pls answer this ..!
.
Prove that n^2 - n is divisible by 2 for any positive integer n .

Answer 1

Casei: Let n be an even positive integer.

When n = 2q

In this case , we have

n2 - n = (2q)2 - 2q = 4q2 - 2q = 2q (2q - 1 )

n2 - n = 2r , where r = q (2q - 1)

n2 - n is divisible by 2 .

Case ii: Let n be an odd positive integer.

When n = 2q + 1

In this case

n2 -n = (2q + 1)2 - (2q + 1)= (2q +1) ( 2q+1 -1)= 2q (2q + 1)

n2 - n = 2r , where r = q (2q + 1)

n2 - n is divisible by 2.

∴ n 2 - n is divisible by 2 for every integer n

Answer 2

n^2-n = n(n-1)
it means its a product of 2 consecutive numbers..
so, one of them is odd and other will be even..
and even numbers are divisible by 2...
hence, n^2-n is divisible by 2...
proved.......