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Math Help - [SOLVED] Need help with this mathematical induction

  1. #1
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    [SOLVED] Need help with this mathematical induction

    This is what I have to prove:

    2(1^3 + 2^3 + ... n^3) = \left[\frac{n(n+1)}{2}\right]^2
    I get the first part. I plugged in 1 and it works fine.

    Now it's the second step that I have trouble with.

    This is what I did:
    1^3 + 2^3 ... + ^3 + (k+1)^3 = \left[\frac{k(k+1)}{2}\right]^2

    Then I added \frac{2(k+1)^3}{2}.

    So now I have:
    \left[\frac{k(k+1)}{2}\right]^2 + \frac{2(k+1)^3}{2}

    How do I go about to simplifying this.

    Do I square the denominator on the first equation and pull out a (k+1)^2 ? Then put the second equation over 4?

    Thank you !
    Last edited by Mrs. White; September 28th 2009 at 08:12 PM. Reason: Better encoding
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  2. #2
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    Talking

    I'm not quite sure what you're doing...? It almost looks like you've assumed the k+1 step, and then did something to it, like working backwards...?

    First, you need to show that the equation is true for n = 1.

    Once you've done that, assume that the equation is true for some n = k:

    2(1^3\, +\, 2^3\, +\, ...\, +\, k^3)\, =\, \left[\frac{k(k\, +\, 1)}{2}\right]^2

    You then do the k+1-th step for the left-hand side, and see where that leads:

    2(1^3\, +\, 2^3\, +\, ...\, +\, k^3\, +\, (k\, +\, 1)^3)

    2(1^3\, +\, 2^3\, +\, ...\, \, k^3)\, +\, 2(k\, +\, 1)^3

    Now substitute from the assumption step:

    \left[\frac{k(k\, +\, 1)}{2}\right]^2\, +\, 2(k\, +\, 1)^3

    Convert to a common denominator, combine the terms, and see what you get. It might help to note that you're wanting eventually to end up with:

    \left[\frac{(k\, +\, 1)(k\, +\, 2)}{2}\right]^2

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  3. #3
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    Quote Originally Posted by stapel View Post
    I'm not quite sure what you're doing...? It almost looks like you've assumed the k+1 step, and then did something to it, like working backwards...?

    First, you need to show that the equation is true for n = 1.

    Once you've done that, assume that the equation is true for some n = k:

    2(1^3\, +\, 2^3\, +\, ...\, +\, k^3)\, =\, \left[\frac{k(k\, +\, 1)}{2}\right]^2

    You then do the k+1-th step for the left-hand side, and see where that leads:

    2(1^3\, +\, 2^3\, +\, ...\, +\, k^3\, +\, (k\, +\, 1)^3)

    2(1^3\, +\, 2^3\, +\, ...\, \, k^3)\, +\, 2(k\, +\, 1)^3

    Now substitute from the assumption step:

    \left[\frac{k(k\, +\, 1)}{2}\right]^2\, +\, 2(k\, +\, 1)^3

    Convert to a common denominator, combine the terms, and see what you get. It might help to note that you're wanting eventually to end up with:

    \left[\frac{(k\, +\, 1)(k\, +\, 2)}{2}\right]^2

    I'm not quite sure what I'm doing myself. This subject was not covered in class, but I'm doing an extra assignment and this is a question.

    I'm trying to follow the instructions in the book but they are not very clear.

    So now if I put both terms on same denominator I end up with

    \left[\frac{k(k+1)}{2}\right]^2 + \frac{8(k+1)^3}{4} ?


    Should it be: \frac{(k+1)^2 (k) + 8(k+1)}{4} ?

    I appreciate your help greatly. I have spent my whole weekend researching these and being stressed out so getting some help is appreciated.
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  4. #4
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    Ok so I reworked the problem.


    From
    \left[\frac{k(k+1)}{2}\right]^2 + (k+1)^3

    I got this

    \frac{(k^2+k)^2}{4} + (k^3+3K^2+3k+1)

    Then I got:

    \frac{(k^4+2k^3 +k^2)}{4} + \frac{4(k^3+3K^2+3k+1)}{4}

    Now when I combine like terms I got

    \frac{k^4+6k^3+13k^2+12k+4}{4}

    And now I'm stuck again. I want to try to factor this out. I'll work some more on it...
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  5. #5
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    I got it ! Thanks for the help.
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