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Math Help - Using mathematical induction...

  1. #1
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    Using mathematical induction...

    n
    I'm asked to show that Σ k^3 = [(n(n+1))/2]^2, ∀n ∈ ℕ
    k=1
    Can anyone help?
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  2. #2
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    Quote Originally Posted by Lindsaijoyce View Post
    n
    I'm asked to show that Σ k^3 = [(n(n+1))/2]^2, ∀n ∈ ℕ
    k=1
    Can anyone help?
    Hi Lindsai,

    P(k)

    \sum_{} k^3=\left(\frac{k(k+1)}{2}\right)^2

    P(k+1)

    \sum_{} (k+1)^3=\left(\frac{(k+1)(k+2)}{2}\right)^2

    Test if P(k) true causes P(k+1) true

    Proof

    If P(k) is true, then the sum of k+1 cubes is

    \sum_{} k^3+(k+1)^3=\left(\frac{k(k+1)}{2}\right)^2+(k+1)^  3

    Now you only need verify that this is what you get when you expand P(k+1).

    Then test the formula for the very first cube.


    \frac{(k+1)(k+1)}{2}\ \frac{(k+2)(k+2)}{2}=\frac{(k+2)(k+2)(k+1)^2}{2^2}

    =\frac{\left(k^2+4k+4\right)(k+1)^2}{2^2}=\left(\f  rac{k(k+1)}{2}\right)^2+4(k+1)\left(\frac{k+1}{2}\  right)^2

    =\left(\frac{k(k+1)}{2}\right)^2+(k+1)^3

    test for n=1

    1^3=\left(\frac{1(1+1)}{2}\right)^2
    Last edited by Archie Meade; June 11th 2010 at 04:01 AM.
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  3. #3
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    Just to mention something:

    she should do the base inductive step at the start. That is check if it works for n=1.
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  4. #4
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    Quote Originally Posted by p0oint View Post
    Just to mention something:

    she should do the base inductive step at the start. That is check if it works for n=1.
    What????

    Does she have to do it the

    1. test for the first term
    2. test for n+1

    way ?

    that's just a format for anyone who wants an answer without knowing much about
    induction.

    As I do not advocate that, I do't do it that way.

    You're welcome to of course!

    The inductive step discovers a term-by-term relationship.

    It finds out if the following relationship exists....

    true for k=1 causes true for k=2
    true for k=2 causes true for k=3
    true for k=3 causes true for k=4

    to infinity.

    It's like stacking up an endless sequence of dominoes.

    Then, if you topple the first domino, they all fall.

    Testing for "true for the 1st term" at that stage then proves the formula is true to infinity.

    To say you must prove for T1 first and then do the inductive step
    and declaring that that's the way only presents a compressed dumbed-down version.
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  5. #5
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    Quote Originally Posted by p0oint View Post
    Just to mention something:

    she should do the base inductive step at the start. That is check if it works for n=1.
    No, most people do it that way because that is the simplest part, but there is no requirement that one be done before the other.
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