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Finding the sum of 1^3 + 2^3 + + n^3 by induction

  1. Jul 6, 2014 #1
    [tex]1^3+2^3+...+n^3 = \left[ \frac{n(n+1)}{2}\right]^2; n\ge 1[/tex]

    [tex]P(1) = 1^3 = \frac{8}{8} = 1[/tex]

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

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

    I start getting stuck here
    I foiled it out then let m = P(k)

    [tex]\left[ m + \frac{2(k+1)}{2}\right]^2[/tex]
    Last edited: Jul 6, 2014
  2. jcsd
  3. Jul 6, 2014 #2
    You probably meant ##m = \sqrt{P(k)}##. Now expand the square and use the induction hypothesis.
  4. Jul 6, 2014 #3


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    You assume that [tex]P(k) = 1^3+...+k^3 = \left[ \frac{k(k+1)}{2}\right]^2[/tex] To get P(k+1), you have to add (k+1)^3 to P(k). [tex]P(k+1)=P(k)+(k+1)^3=\left[ \frac{k(k+1)}{2}\right]^2+(k+1)^3[/tex] Write it out, and show that it is equal to [tex]P(k+1) = \left[ \frac{(k+1)(k+2)}{2}\right]^2[/tex]
    pull out (k+1)2

  5. Jul 6, 2014 #4

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

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

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

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

  6. Jul 6, 2014 #5


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    Apply that a2b2=(ab)2

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

    is it familiar? Is it what you wanted to arrive at?

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