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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)

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  1. anonymous
    • 5 years ago
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    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

  2. anonymous
    • 5 years ago
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    fan me if you understand :)

  3. anonymous
    • 5 years ago
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    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.

  4. anonymous
    • 5 years ago
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    Thank you very much :)

  5. anonymous
    • 5 years ago
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    indeed i was wrong :) but its great u understand :)

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