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Proof of a Sequence

  1. Aug 23, 2004 #1
    Hello all

    I encountered the following problem in my textbook

    Prove (1^3 + 2^3 + ... + n^3 ) = ( 1 + 2 +... + n) ^2.

    Here is my solution, however I become stuck on the final part of the problem.

    Let S1 = 1/2(n+1)

    Let S2 = 1/6n(n+1)(2n+1)

    S3 = (1^3 + 2^3 + ... + n^3)

    (1^3 + 2^3 + ... + n^3 ) = (v+1)^4 - v^4 = (v^4 + 4v^3 + 6v^2 + 4v + 1) - v^4

    = 4v^3 + 6v^2 + 4v + 1.

    Substituting v = 0 , 1 , 2, ... , n into the equation and adding we get

    4(0)^3 + 6(0)^2 + 4(0) + 1​
    4(1)^3 + 6(1)^2 + 4(1) + 1​
    4(2)^3 + 6(2)^2 + 4(2) +1​
    4(n)^3 + 6(n)^2 + 4(n) + 1​


    = 4S3 + 6S2 + 4S1 + n + 1

    (n+1) ^4 = 4S3 + 6S2 + 4S1 + n + 1

    Since we are solving for S3, we collect terms and get

    4S3 = (n+1) ^4 - 6S2 - 4S1 - n - 1

    = (n+1) ([n+1]^3 - 1- n(2n+1) - 2n)

    = (n+1) (n^3 + 3n^2 +3n + 1) - (1 + n(2n+1) +2n)

    = (n+1) (n^3 + 3n^2 +3n +1 ) - (1 + 2n^2 +n +2n)

    = (n+1) (n^3 + n^2 )

    S3 = ¼ (n+1)(n^3 + n^2)

    Here is where I become stuck.

    How does S3 = ( 1 + 2 +... + n) ^2?

    Any help would be greatly appreciated.

    Thanks

    Also do you know where I can get LaTex?
     
    Last edited: Aug 23, 2004
  2. jcsd
  3. Aug 23, 2004 #2

    arildno

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    [tex]S3=\frac{n^{2}(n+1)^{2}}{2^{2}}=(\frac{n(n+1)}{2})^{2}=(\sum_{i=1}^{n}i)^{2}[/tex]

    Hope that helps..:wink:
     
  4. Aug 23, 2004 #3
    Have you tried proof through induction? Thats the first thing that always comes to my mind when I see these types of problems.
     
  5. Aug 23, 2004 #4
    Thanks a lot arildno. That really helped me!
     
  6. Aug 23, 2004 #5
    This exact same problem occurred in one of my final exams. The question required a poof by induction.

    -Ray.
     
  7. Aug 23, 2004 #6
    Are you asking how to use LaTeX in these forums? One fast way to obtain the reference PDF is by clicking directly on arildno's formatted text. This will open a window that will contain a link.

    -Ray.
     
  8. Aug 24, 2004 #7
    how would i use the formula for 1^2 + 2^2 +... +n^2 = 1/6n(n+1)(2n+1) to find a formula for

    1^2 + 3^2 +... (2n+1)^2. Would i use the same method i used in my other problem?

    Any help would be appreciated.

    Thanks!
     
  9. Aug 24, 2004 #8

    HallsofIvy

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    1^2 + 3^2 +... (2n+1)^2 (sum of squares of odd numbers) is 1^2 + 2^2 +... +n^2- 2^2+ 4^2+ 6^2+ ... (sum of all squares minus sum of square of even numbers) and, of course, 2^2+ 4^2+ 6^2+... is 4(1^2+ 2^2+ 3^2+...).

    Be careful about the limits! In order to find 1+ 3^2+ 5^2+ 7^2 (i.e. n= 3 since 7= 2(3)+1 you would find 1^2+ 2^2+ 3^2+ 4^2+ 6^2+ 7^2+ 8^2 - (2^2+ 4^2+ 6^2+ 8^2)
    = 1^2+ 2^2+ 3^2+ 6^2+ 7^2+ 8^2- 4(1^2+ 2^2+ 3^2+ 4^2). You would then use the formula for sums of square with n= 8 and n= 4. In general, if we let S(n) mean "the sum of squares up to n^2" (the original formula), the sum of odds up to 2n+1 would be S(2n+2)- 4S(n+1).
     
  10. Aug 24, 2004 #9
    Thanks a lot. That really helped!
     
  11. Aug 24, 2004 #10
    HallsofIvy, is that just a typo or did you make a mistake in counting 5^2 and 4^2?
     
  12. Aug 24, 2004 #11

    HallsofIvy

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    Yeah, I dropped 5^2 in one sum and both 5^2 and 4^2 in another. Thanks for catching that.
     
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