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Sum of cubes - please verify my proof

  1. Mar 17, 2009 #1
    1. The problem statement, all variables and given/known data
    Prove

    [tex]\sum^{2n}_{r=1} r^3 = n^2(2n+1)^2[/tex]

    by mathematical induction.

    My proof is below but it is rather long winded. Is there a quicker way of doing the algebraic manipulation? Preferably one that forgoes the polynomial division and multiplying out all the brackets!

    3. The attempt at a solution
    Base case, [tex]n=1[/tex]:

    [tex]\sum^{2 \times 1}_{r=1} r^3 = 1^2 (2 \times 1+1)^2 = 1 \times 9 = 9[/tex]

    [tex]= 1^3 + 2^3 = 1 + 8 = 9[/tex]

    Inductive step, assume true for [tex]n=k, \quad k \in \mathbb{N}[/tex]

    [tex]\sum^{2k}_{r=1} r^3 = k^2(2k+1)^2[/tex]

    Adding the next two terms, [tex]n=k+1[/tex]

    [tex]\sum^{2(k+1)}_{r=1} r^3 = k^2(2k+1)^2 + (2k+1)^3 + (2k+2)^3[/tex]

    Multiplying out and collecting terms
    [tex]\sum^{2k+2}_{r=1} r^3 = 4k^4 + 20k^3 + 37k^2 + 30k + 9[/tex]

    By factor theorem, (k+1) is a factor.
    By long-hand polynomial division

    [tex]= (k+1)(4k^3 + 16k^2 + 21k + 9)[/tex]

    By factor theorem, (2k+3) is a factor.
    By long-hand polynomial division

    [tex]= (k+1)(2k+3)(2k^2 + 5k + 3)[/tex]

    Factorising the quadratic
    [tex]= (k+1)(2k+3)(2k+3)(k+1)[/tex]
    [tex]= (k+1)^2(2k+3)^2[/tex]
    [tex]= (k+1)^2(2(k+1)+1)^2[/tex]

    So if the statement is true for [tex]n=k[/tex], it is true for [tex]n=k+1[/tex].
    Therefore
    [tex]\sum^{2n}_{r=1} r^3 = n^2(2n+1)^2[/tex]
    for all [tex] n \geq 1, \quad n \in \mathbb{N}[/tex]

    Thanks for any suggestions on making this simpler.
     
  2. jcsd
  3. Mar 18, 2009 #2

    lanedance

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    Homework Helper

    Hi iainfs

    how about this

    [tex]\sum^{2(k+1)}_{r=1} r^3 = k^2(2k+1)^2 + (2k+1)^3 + (2k+2)^3[/tex]

    Then re-write everything in terms of k+1

    [tex]\sum^{2(k+1)}_{r=1} r^3 = ((k+1)-1)^2(2(k+1)-1)^2 + (2(k+1)-1)^3 + (2(k+1))^3[/tex]

    then to make the algebra simpler change variable to m = k+1
    [tex]\sum^{2(k+1)}_{r=1} r^3 =\sum^{2m}_{r=1} r^3 = (m-1)^2(2m-1)^2 + (2m-1)^3 + (2m)^3[/tex]

    and then group terms and see where we end up
    [tex]\sum^{2m}_{r=1} r^3 = ((m-1)^2 + (2m-1))(2m-1)^2 + (2m)^3[/tex]
    [tex]= m^2(2m-1)^2 + (8m)m^2[/tex]
    [tex]= m^2((2m-1)^2 + 8m)[/tex]
    [tex]= m^2((4m^2 -4m +1) + 8m)[/tex]
    [tex]= m^2(4m^2 +4m +1)[/tex]
    [tex]= m^2(2m +1)^2[/tex]
    ...sweet

    I always find that variable change always makes it easier for me...
     
  4. Mar 18, 2009 #3
    Thank you! Knew there would be a more elegant way.
     
  5. Mar 19, 2009 #4
    Doesn't this work for all 2n in N too? (not just n in N) for 2n>=1.
    Because adding (2n+1)^3 to both sides gives,
    Sum(from r=1 to 2n+1) of r^3
    = n^2*(2n+1)^2+(2n+1)^3
    =(2n+1)^2*(n+1)^2
    =((2n+1)/2))^2+((2n+1)+1)^2
    and the initial cases of 2n=1, 2n=2, 2n=3 all work.
     
  6. Mar 19, 2009 #5

    lanedance

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    Homework Helper

    n is an integer, so not too sure what you mean by 2n=3
     
  7. Mar 22, 2009 #6
    no i'm saying that it works for all 2n in N. so n does not have to be an integer for this to work- n can be 0.5, 1, 1.5, 2,...
     
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