Solution: Since the set starts with an odd number (1) and has an odd number of integers, the set would end with an odd number, too.

Let’s see the set. Set N: {1, 2, 3, 4, 5, ..., (2n + 1)}, where n is a positive integer.

=> Number of odd terms is one more than the number of even terms.

Thus, the number of odd terms = (n + 1) and we derive a new set from set N, that is {1,3,…, 2n+1}.

Then we get the average of the odd integers = \(\frac{\left[1\ +\ 3\ +\ ......\ +\ \left(2n\ +\ 1\right)\right]}{\left(n\ +\ 1\right)}\) and since \(1\ +\ 3\ +\ ......\ +\ \left(2n\ +\ 1\right)\ =\ \left(n\ +\ 1\right)^2\), we get \(\frac{\left[1\ +\ 3\ +\ ......\ +\ \left(2n\ +\ 1\right)\right]}{\left(n\ +\ 1\right)}=\frac{\left(n\ +1\right)^2}{\left(n\ +\ 1\right)}\ =\ n\ +\ 1\)

The number of even terms = n and we derive a new set from set N, that is {2,4,…, 2n}.

Then we get the average of the even integers = \(\frac{\left(2\ +\ 4\ +\ ....\ +\ 2n\right)}{n}=\frac{2\cdot\left(1\ +\ 2\ +\ ....\ +\ n\right)}{n}\) and since \(1\ +\ 2\ +\ ........\ +\ n\ =\ \frac{n\left(n\ +\ 1\right)}{2}\), we get \(\frac{\left(2\ +\ 4\ +\ .......\ +\ 2n\right)}{n}=\frac{2\cdot\left(1\ +\ 2\ +\ .......\ +\ n\right)}{n}=\frac{n\left(n\ +\ 1\right)}{n}\ =\ n\ +1\)

Therefore, we get the difference of the average of the odd integers and the average of the even integers in set N=(n + 1) - (n + 1) = 0

Say there are only three terms in the set

S = {1, 2, 3}

=> X = \(\frac{\left(1\ +\ 3\right)}{2}=\frac{4}{2}=2\)

=> Y = 2

=> X – Y = 0

Again, say there are five terms in the set

S = {1, 2, 3, 4, 5}

=> X = \(\frac{\left(1\ +\ 3\ +\ 5\right)}{3}=\frac{9}{3}=3\)

=> Y = \(\frac{\left(2\ +\ 4\right)}{2}=\frac{6}{2}=3\)

=> X – Y = 0

Therefore, C is the correct answer.

Answer C