1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    [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

    User Avatar
    Staff Emeritus
    Science Advisor

    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

    User Avatar
    Staff Emeritus
    Science Advisor

    Yeah, I dropped 5^2 in one sum and both 5^2 and 4^2 in another. Thanks for catching that.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Proof of a Sequence
  1. Proof of a Sequence (Replies: 7)

  2. Proof of sequence (Replies: 2)

  3. Sequence Proof (Replies: 8)

Loading...