The idea of the above is right, but the numbers at the end aren't correct.
The first n positive even integers are 2, 4, 6, 8, ..., 2n.
Those are the first n positive even integers. Note that the last one is 2n.
So h = 2 + 4 + 6 + ... + 2n.
The first 3 odd positive integers are 1, 3, 5.
The first 5 odd positive integers are 1, 3, 5, 7, 9.
The first 6 odd positive integers are 1, 3, 5, 7, 9, 11.
What is the list of the first n odd positive integers? What does it end at?
Can you see the pattern?
If you list the first three, it ends at 2(3) - 1.
If you list the first five, it ends at 2(5) - 1.
...
If you list the first n, it ends at 2n - 1.
So then k = 1 + 3 + 5 + ... + (2n - 1).
Then h - k = 1 + 1 + ... + 1 as suggested above, and there are n ones since there are n terms.
So h - k = n.