Question

Prove that n²-n is divisible by 2 for every positive integer n.

Answer 1

n = 2 q +r ; r = 0 ,1

if r = 0

(2q)^2 -2q
4q^2 -2q
2q (q-1)
similarly r = 1 simplyfy them

Answer 2

Hii friend,

We know that any positive integer is in the form of 2Q or 2Q+1 , for some integer Q.

Now, we have two cases


Case(1) when n = 2Q

In this case , we have

n²-n = (2Q)² - 2Q

=> 4Q²-2Q = 2Q(2Q-1) = r , where r = Q(2Q-1) is an integer.

=> n²-n is divisible by 2.

Case(2)
When n = 2Q+1

Here we have,

n²-n = (2Q+1)² - (2Q+1)

=> (2Q+1) (2Q+1-1)

=> 2Q(2Q+1) = 2r , where r = Q(2Q+1) is an integer.

Therefore,

n²-n is divisible by 2

HENCE,

n²-n is divisible by 2 for every positive integer n.


HOPE IT WILL HELP YOU...... :-)