A = {(n + 1)2 − n2 | n ∈ N}is the set of all odd numbers.Write down a similar expression for the set, B, of all even numbers.List the elements of the setsC = {(n + 1)2 − (n2 + 2n) | n ∈ N}and (A ∩ B) ∪ C.

First, let's find a similar expression for the set B of all even numbers. We know that the difference between two consecutive squares is an odd number, as given by the expression for set A. To get an even number, we can simply add another 1 to this difference. So, the expression for set B would be: B = {(n + 1)2 - n2 + 1 | n ∈ N}.

Next, let's list the elements of the set C. The expression for set C simplifies to C = {(n + 1)2 - n2 - 2n | n ∈ N} = {1 - 2n | n ∈ N}. For n = 1, 2, 3, ..., the elements of set C would be -1, -3, -5, ..., which are negative odd numbers.

Finally, let's find the elements of the set (A ∩ B) ∪ C. The intersection of sets A and B, A ∩ B, would be an empty set, because A consists of odd numbers and B consists of even numbers, and there are no numbers that are both odd and even. Therefore, (A ∩ B) ∪ C would simply be set C, which is the set of negative odd numbers.