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(.
