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Sum of the first n cubes

  1. Nov 7, 2007 #1
    1. The problem statement, all variables and given/known data
    Can someone help me derive the formula for the sum of the first n cubes from the formula for the sum of the first n integers that elucidates the reason why the former is the square of the latter?


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 7, 2007 #2

    rock.freak667

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    Well you could consider the sum from n=1 to N of (n+1)^4-n^4 and then expand using the method of differences
     
  4. Nov 7, 2007 #3
    You mean use the fact that it telescopes?

    Then you just get N^4-1? How does that help?
     
  5. Nov 8, 2007 #4

    Gib Z

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    [tex](n+1)^4 - n^4 = 4n^3 + 6n^2 + 4n + 1[/tex] by the Binomial theorem.

    Sum BOTH sides, and the LHS should get (N+1)^4 actually. Do you see how the RHS will be expressions in terms of the first N integers to the power of 3, 2, 1 and 0? You already know the cases for 0, 1 and 2, now you can work out 3.
     
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