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anonymous
 5 years ago
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)
anonymous
 5 years ago
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)

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0h=2+4+6+8+10...+n4+n2+n k=1+3+5+7+9...+n5+n3+n hk=(21)+(43)+(65)+(87)+(109)+...+(n4  (n5))+(n2+(n3))+nn hk=1+1+1+1+1...+1+1+1=n

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0fan me if you understand :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Thank you very much :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0indeed i was wrong :) but its great u understand :)
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