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Prove that for each n in N, 1^3+2^3+ +n^3=[n(n+1)/2]^2

  1. Aug 29, 2011 #1


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    prove that for each n in N, 1^3+2^3+...+n^3=[n(n+1)/2]^2
  2. jcsd
  3. Aug 29, 2011 #2
    What did you try?? If you let us know where you're stuck, then we'll know how to help...
  4. Aug 29, 2011 #3


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    my dumb butt got "lost" in the wanting to prove p(k+1) is true (if that even makes sense, lol)
  5. Aug 29, 2011 #4
    and where are you stuck in induction??
  6. Aug 29, 2011 #5


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    the proving of p(k+1) is true.
    basically the area where for each k>=1, if P(k) is true, then p(k+1) is true.....basically the induction part, lol
  7. Aug 29, 2011 #6
    Yes, so to show that p(k+1) is true, you need to prove that


    Now, what happens if you appy "p(k) is true" on that??
  8. Aug 29, 2011 #7
    Induction has a few steps. Let's see if this clarifies them a bit,

    1) Base Case: Show that your summation formula works for k = 1 case (which is probably easiest here lol)

    2) Induction Case: Create an induction hypothesis. For this case, you assume that the kth case holds. In other words,

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

    is true. Now, show that the kth case implies the (k+1)th case. How do you think you can do this?
    Last edited: Aug 29, 2011
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