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Square of the sum = Sum of the cubes

  1. Nov 6, 2009 #1
    (1+2+....n)[tex]^{2}[/tex] = 1[tex]^{3}[/tex]+2[tex]^{3}[/tex]+...n[tex]^{3}[/tex]

    [tex]\frac{n^{2}(n+1)}{4}^{2}[/tex] = 1[tex]^{3}[/tex]+2[tex]^{3}[/tex]+...n[tex]^{3}[/tex]

    how do you simplify the right side to show that they are equal?
     
  2. jcsd
  3. Nov 6, 2009 #2
    A much better idea is to use induction to prove the first equation.

    Petek
     
  4. Nov 13, 2009 #3
    What exactly is the question?
    If the answer is to make you prove that square of sum equal to sum of cubes , you have to use mathematical induction to prove that sum is of the following formula [tex]\frac {n(n+1)}{2}[/tex] and once again use mathematical induction to prove that the sum of cubes is of the following forumula [tex][\frac {n(n+1)}{2}]^{2}[/tex].Then only you can draw your conclusion.
     
  5. Nov 13, 2009 #4

    D H

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    The question obviously is to prove (1+2+...+n)2 = 13+23+...n3.

    The obvious approach, as Petek noted, is to use induction.
     
  6. Nov 14, 2009 #5

    HallsofIvy

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    The standard simplification for the right side is
    [tex]\frac{n^2(n+1)^2}{4}[/tex]
    And, as has been suggested several times now, can be proven by induction on n.
     
  7. Nov 14, 2009 #6
    Hm, could someone please briefly explain induction? This thread has sparked my curiosity. I feel like I should know induction to n, but I don't :(
     
  8. Nov 14, 2009 #7

    CRGreathouse

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    Induction: if you can prove
    1. f(1) is true
    2. If f(n) is true, then f(n+1) is true
    for any proposition f, then f(n) is true for all n.
     
  9. Nov 14, 2009 #8

    D H

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    The basic idea behind using induction to prove some indexed relationship fn is
    • Show that the relationship is true for some particular value of n0.
    • Show that if the relationship is true for n0, n0+1,n0+2, ..., N, then it is true for N+1.
    By induction, these two items mean that the relationship is true for all integers greater than or equal to n0.

    The problem at hand is to show that 1+2+...+n)2 = 13+23+...n3. This equality is trivially true for the case n=1; it reduces to 12=13=1. The trick then is to show that if the relationship is true for some particular n, then it is also true for n+1.
     
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