Question

Show 2 different solutions to the task. Prove that for every integer n (...-3, -2, -1,...

Show 2 different solutions to the task.

Prove that for every integer n (...-3, -2, -1, 0, 1, 2, 3, 4...), the expression n2 + n will always be even.

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Answer #1

Defination : An integer n is said to be even if n = 2k for some integer k and an integer m is said to be odd if m=2k+1 for some integer k .

Suppose n is given integer .

n is either even or odd .

Case 1 : If n is even .

Then n = 2k for some integer k . Now ,

=  

=

=

So is even as it is multiple of 2 .

Case 2 : If n is odd .

Then n = 2k +1 for some integer k . Now ,

=

=

=

=

So is even as it is multiple of 2 .

i.e., n is even or odd is always even .

Hence is even for every integer n .

.

.

.

If you have any doubt or need more clarification at any step please comment.

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