Verified answer

Step-by-step explanation:

Any positive integer is of the form 2q , 2q + 1 , 2q + 2 , where q is some integer

When n = 2q

Then,

n² - n = (2q)² - 2q

4q² - 2q

2q(2q-1)

which is divisible by 2

And

when n = 2q + 1

Then,

n² + n = (2q + 1)² + 1

(2q)² + 2 × 2q × 1 + 1 + 1

4q² + 4q + 2

2(2q²+2q+1)

which is divisible by 2

And

when n = 2q+2

Then,

n² + n = (2q+2)² + 2

(2q)² + (2)² + 2 × 2q × 2 + 2

4q² + 4 + 8q + 2

4q² + 8q + 6

2(2q² + 4q + 3)

which is divisible by 2.

Hence,

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

Step-by-step explanation:

Let us take n in the form of 2m for even or 2m+1for odd where m is any +ve no.

so for 2m

n^2-n=n(n-1)=2m(2m-1)-------» even and divisible by 2

for n=2m+1

n^2-n=n(n-1)=(2m+1)(2m)=2a ------> where a= m(2m+1)

and divisible by 2

therefore prooved