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Math Help - 1^3 +....+n^3 =... (induction)

  1. #1
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    1^3 +....+n^3 =... (induction)

    1^3+2^3+...+n^3 = \left[ \frac{n(n+1)}{2}\right]^2; n\ge 1

    P(1) = 1^3 = \frac{8}{8} = 1

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

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

    from here I foiled and let m = p(k) for

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

    \left[ m+k+1\right]^2
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  2. #2
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    Re: 1^3 +....+n^3 =... (induction)

    Quote Originally Posted by Jonroberts74 View Post
    1^3+2^3+...+n^3 = \left[ \frac{n(n+1)}{2}\right]^2; n\ge 1

    P(1) = 1^3 = \frac{8}{8} = 1

    P(k) = 1^3+...+k^3 = \left[ \frac{k(k+1)}{2}\right]^2
    Okay, you want to prove that if P(k) is true then so is P(k+1).

    P(k+1) = 1^3+...+k^3+(k+1)^3 = \left[\frac{(k+1)(k+2)}{2}\right]^2
    No, you can't say this. You don't know it is true, this what you want to prove
    What you can say is that 1^3+ ...+ k^3+ (k+ 1)^3= \left(1^3+ ...+ k^3\right)+ (k+1)^3= \left[\frac{k(k+1)}{2}\right]^2+ (k+1)^3

    Now try to show that the right side is the same as \left[\frac{(k+1)(k+2)}{2}\right]^2. For example you can factor (k+1)^2 out of both terms.

    from here I foiled and let m = p(k) for

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

    \left[ m+k+1\right]^2
    Thanks from topsquark
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  3. #3
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    Re: 1^3 +....+n^3 =... (induction)

    1^3 +....+n^3 =... (induction)-sum-cubes.jpg
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    Re: 1^3 +....+n^3 =... (induction)

    I thought you assume P(k) to be true then show P(k+1) is true therefore P(k) is also true

    I also can't see what the image is, even opening it it is too small for me. I can only read the title of it, I googled it but couldn't find the same image
    Last edited by Jonroberts74; July 5th 2014 at 10:24 PM.
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    Re: 1^3 +....+n^3 =... (induction)

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

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

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

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

    ??
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