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Math Help - (Series) Prove the following:

  1. #1
    Super Member Quacky's Avatar
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    (Series) Prove the following:

    Show that when n is even,

    <br />
1^3-2^3+3^3-...-n^3 \equiv 1^3 + 2^3 + 3^3 +...+ n^3 - 16(1^3 + 2^3 + 3^3 +...+(\frac{n}{2})^3) \equiv \sum\limits_{r=1}^{n} r^3 - 16 \sum\limits_{r=1}^{\frac{n}{2}} r^3

    I don't know how to approach this question, the 'when n is even' is throwing me, and even without that part, I think I'd struggle. How should I begin?
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  2. #2
    Member HappyJoe's Avatar
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    Did you try proving it by induction? It seems fit for this kind of problem.
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  3. #3
    Super Member Quacky's Avatar
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    No, I didn't try induction. The reason is that this chapter is before induction in the textbook, and we didn't cover induction in class untill after the work was set. I suppose it's worth a try though.
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  4. #4
    Member Traveller's Avatar
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    Did you cover the formula for summing cubes ?
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  5. #5
    Super Member Quacky's Avatar
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    Yes Traveller, if by that you mean :

    \sum\limits_{r=1}^{n} r^3 = \frac{n^2}{4}(n+1)^2

    But I still can't see a way of applying it to this question specifically.
    Last edited by Quacky; September 29th 2010 at 11:14 AM.
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  6. #6
    Member Traveller's Avatar
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    Quote Originally Posted by Quacky View Post
    Yes traveller, if by that you mean :

    \sum\limits_{r=1}^{n} r^3 = \frac{n^2}{4}(n+1)^2

    But I still can't see a way of applying it to this question specifically.
    You can write each odd number as 2k-1 and each even number as 2k and then apply this formula. Then you can add and subtract suitable terms to bring it to the required form.
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  7. #7
    Super Member Quacky's Avatar
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    I'll try that now.
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  8. #8
    Super Member Quacky's Avatar
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    It seems I am still unable to solve this problem. Would you elaborate on your post above slightly? I've clearly misunderstood your method.
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  9. #9
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    1- 2^3+ 3^3- 4^3+ \cdot\cdot\cdot+ (n-1)^3+ n^3= (1+ 2^3+ 3^3+ 4^3+ \cdot\cdot\cdot+ n^3)- 2(2^3+ 4^3+ 6^3+ n^3)
    for n even.
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  10. #10
    Member Traveller's Avatar
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    Quote Originally Posted by Quacky View Post
    It seems I am still unable to solve this problem. Would you elaborate on your post above slightly? I've clearly misunderstood your method.

    Sorry, I was wrong.

    HallsofIvy is correct. You don't need anything else to prove the formula.
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  11. #11
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    Hello Quacky,
    Could you please tell me the name of the book from which you got the sum you posted above ? Thanks a lot .....
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  12. #12
    Super Member Quacky's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    1- 2^3+ 3^3- 4^3+ \cdot\cdot\cdot+ (n-1)^3+ n^3= (1+ 2^3+ 3^3+ 4^3+ \cdot\cdot\cdot+ n^3)- 2(2^3+ 4^3+ 6^3+ n^3)
    for n even.
    Yeah, I see this now! From here, I simply have to take out a factor of 2^3 from the last part of the second expression, which gives:

     (1+ 2^3+ 3^3+ 4^3+ \cdot\cdot\cdot+ n^3)- 2(2^3)(1^3+ 2^3+ 3^3+ (\frac{n}{2})^3)

    Which can then be summarised by the third expression which I am asked to show in the question.

    Thanks for the helpful reply.
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