Question

The sum of the first n even positive integers is h. The sum of the first n odd positive integers is k. Then h - k = ?
What would I do to solve this? How do I prove it? (the answer is: n)

Answer 1

h=2+4+6+8+10...+n-4+n-2+n
k=1+3+5+7+9...+n-5+n-3+n
h-k=(2-1)+(4-3)+(6-5)+(8-7)+(10-9)+...+(n-4 - (n-5))+(n-2+(n-3))+n-n
h-k=1+1+1+1+1...+1+1+1=n

Answer 2

fan me if you understand :)

Answer 3

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.

Answer 4

Thank you very much :)

Answer 5

indeed i was wrong :) but its great u understand :)