Question

True or false. If a is any odd integer, then a^2 + a is even. Explain this.

Answer 1

Answer:

True.

Step-by-step explanation:

We can represent an odd number by 2n + 1 where n = 0, 1, 2, 3, 5 etc.

Substituting:

a^2 + a = (2n + 1)^2 + 2n + 1

=  4n^2 + 4n + 1 + 2n + 1

= 4n^2 + 6n + 2

= 2(2n^2 + 3n + 1)

which is even because any integer multiplied by an even number is even.

This is also true if we use a negative odd integer:

We have 4n^2 + 4n + 1  - 1 - 2n

= 4n^2 + 2n

=  2(2n^2 + n(.