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Series Solution Check

  1. Sep 11, 2009 #1
    1. The problem statement, all variables and given/known data

    Evaluate the sum [tex]2009[/tex][tex]^{2}[/tex] - [tex]2008[/tex][tex]^{2}[/tex] + [tex]2007[/tex][tex]^{2}[/tex] - [tex]2006[/tex][tex]^{2}[/tex] + ... + [tex]3[/tex][tex]^{2}[/tex] - [tex]2[/tex][tex]^{2}[/tex] + [tex]1[/tex][tex]^{2}[/tex]

    2. Relevant equations

    I think that the equivalent series representation of this sum is:

    [tex]\sum^{2009}_{n=1}n^{2}(-1)^{n+1}[/tex]

    3. The attempt at a solution

    I vaguely remember in one of my calculus classes way back when something about finding the convergence of a series, I just don't remember how exactly to do it. I'm sure that is probably the easiest method to obtaining the sum. I found the sum by a somewhat roundabout method, which just consisted of finding patterns within patterns eventually reducing the 2009-part summation to a 12-part summation as follows:

    435 + 21510 + 53910 + 86310 + 118710 + 151110 + 183510 +215910 + 248310 + 280710 + 313110 + 345510

    resulting in a sum of 2,019,045.

    Could someone who knows how to find the solution to the series please check my answer? Thanks!
     
    Last edited: Sep 11, 2009
  2. jcsd
  3. Sep 11, 2009 #2

    Mark44

    Staff: Mentor

    Your closed-form series representation is incorrect. You have
    Notice that every term in the original problem is squared.

    How did you get this?
    Is this based on the first representation? It might be right, but as you haven't explained how you got it, I have no way of telling whether it's correct short of figuring this out myself, which I am reluctant to do.

    I don't believe the answer you got, 2,019,045, is correct. The 2009 term alone is larger than that by a factor of about 2.
     
  4. Sep 11, 2009 #3
    Ah yes, thanks, forgot to put that squared term in there. I changed it in my original post, is the series representation correct now?
     
  5. Sep 11, 2009 #4
    The way that I reduced it took a few steps.

    First, I summed together adjacent terms to get positive numbers. This reduced the 2009+1-part summation(including the term 0^2 to make the number divisible by 2) to a (2009+1)/2 = 1005-part summation.

    0^2 + 1^2 = 1
    -(2^2) + 3^2 = 5
    -(4^2) + 5^2 = 9
    -(6^2) + 7^2 = 13
    -(8^2) + 9^2 = 17

    Notice how each term is just increasing by 4. So I rewrote this sequence to see if I could find another pattern to again reduce the number of terms to sum.

    1,5,9,13,17,21,25,29,33,37,41,45,49,53,57,61,65,69,73,77

    Notice how the first 5 terms sum is 45. The next 5 terms sum is 145. The next 5 terms sum is 245. The next 5 terms sum is 345...etc. This reduces the previous 1005-part summation to a (1005/5)=201-part summation. I again rewrote this sequence to see if I could find yet another pattern.

    I kept using this method to finally reduce it down to a 12-part summation that I listed I my original post. It seems to make sense to me, but I'm really not sure which is why I was looking for a more direct method so I could check my answer.
     
  6. Sep 11, 2009 #5

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    This doesn't have anything to do with convergence. It is just a sum of finitely many terms. Try grouping it like this:

    [tex] 1^2 + (-2^2 + 3^2) + (-4^2 + 5^2) ... (-2008^2+2009^2)

    = 1+\sum_{k=2}^{2008}(-k^2+(k+1)^2) = 1+\sum_{k=2}^{2008}(2k+1)[/tex]

    The last expression is just an arithmetic series which you can sum. I didn't get your answer but it's late here so I'll let you take it from there.
     
  7. Sep 11, 2009 #6

    Mark44

    Staff: Mentor

    OK, now I understand, and your approach is good one, especially the part where you noticed that each successive pair of terms increases by 4. I get the same value you did for this series, 2,019,045.

    Here is another explanation that uses your idea.

    Let S = (-02 + 12) - (22) + 32) -+ ... - (20082 + 20092)

    There are 1005 pairs of terms, which is easy to see if you index them starting from 0 and ending with 1004.

    So,
    [tex]S~=~\sum_{n = 0}^{1004} (1 + 4n)[/tex]
    This gives is 1 + 5 + 9 + ...
    We can evaluate the summation above by breaking it up into two summations.
    [tex]S~=~\sum_{n = 0}^{1004} 1 + \sum_{n = 0}^{1004}4n[/tex]
    [tex]=~\sum_{n = 0}^{1004} 1 + 4\sum_{n = 1}^{1004}n[/tex]
    (Note that I brought the constant factor outside the 2nd summation and changed the lower limit on the same summation, which doesn't change the final value.)
    = 1005 + 4*1005*502
    = 2,019,045
     
  8. Sep 11, 2009 #7
    Thank you both very much for your help! :biggrin:
     
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