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A series question with r^2

  1. Oct 19, 2004 #1
    A series question with r^2~~

    Hello, all,

    I know this is quite annoying to post two questions within the same week, but I must say this is the best method I can think of when all possibilities in my brain ran out...

    So, this time the question is about series. I have learned it last week in the Mathematical Methods course, the only two clues teacher left me is the formulae of determining the sum of standard arithematic progression and geometric progression. Anyway, in the problem sheet, I am being asked to find a general formula of the [tex] \sum {r^2} [/tex]~~ and also [tex] \sum {r^3} [/tex] as well....

    I have tried the two model methods about A.P. and G.P., but it just cannot simplifies the question down.

    Maybe this is just a question asks me to try all the possibilities, :uhh: I will try them, but also will expects all sorts of help from the forum as well. :redface:

    All right, cheers~
    Last edited: Oct 20, 2004
  2. jcsd
  3. Oct 19, 2004 #2


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    Do you mean the sum : [itex]1^2 + 2^2 + 3^2 + ...+n^2 [/itex] ?

    It's not hard to see (from the first 3 terms) that it is neither an AP nor a GP.

    The sum to n terms is given by [itex]S(n) = \frac{n(n+1)(2n+1)}{6} [/itex]

    I know of no simple, intuitive way to derive this (other than to assume that the sum is a polynomial of order 3, and find its coefficients).
  4. Oct 19, 2004 #3
    Thanks a lot! I have to say this is a great hint you gave me, though I still can't get [itex]S(n) = \frac{n(n+1)(2n+1)}{6} [/itex]. Well, I opened your answer to [itex]\frac{n^3+3n^2+n}{6}[/itex] and found it is obviously an 3rd degree polynomial, and then I started to think the 3rd degrees...

    I find a combained way from series and algebra:
    [itex]\sum(1~n+1) r^3 = 1^3 + 2^3 + 3^3 + ... + n^3 [/itex]
    And if thinking them individually,
    then substitute them into the first equation, we have:
    [itex]\sum r^3 = \sum(0~n) r^3 + \sum(0~n) r^2 + \sum(0~n) r^3 + n[/itex]

    And at this point, I believe I could sort it out, as I know [itex] \sum(0~n) r[/itex] can be substited by if the same method is applied to [itex] \sum(1~n+1) r^2 [/itex]...

    That is really not an intuitively determinable question...
  5. Oct 20, 2004 #4


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    This is quite nice...it'll work nicely. Never thought of something like that before...simple and elegant.

    I think this is what you mean :
    [tex](n+1)^3 = n^3 + 3n^2 + 3n + 1[/tex]
    [tex](n+1)^3 - n^3 = 3n^2 + 3n + 1[/tex]
    [tex]\sum _{n=1} ^m((n+1)^3 - n^3) = \sum _{n=1} ^m (3n^2 + 3n + 1)[/tex]
    [tex](N+1)^3 - 1^3 = 3\sum _{n=1} ^m n^2 + 3\sum _{n=1} ^mn + m[/tex]

    [itex]\sum _{n=1} ^ m n = m(m+1)/2[/itex], is easy to prove geometrically, or by pairing up the numbers as (1,m), (2,m-1), (3,m-2),...each of which has the sum m+1.

    Plugging this result above gives the required result for the sum of squares.
  6. Oct 20, 2004 #5
    the proof he gave is a textbook proof (at least from my book, and it's a high school book).
    the book also gives an explanation how to solve for 1^3+2^3+....+n^3 (you need to use the identity of (a+b)^4), i wonder if someone have a general proof that every time you need to get to the sum equation of the n degree sequence you need to use the identity of (a+b)^(n+1).
  7. Oct 20, 2004 #6
    btw thomasjoe are you a quantum particle? :rolleyes:

    (im asking because it seem you are both in beijing and london in the same time.).
  8. Oct 20, 2004 #7


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    Ouch ! I knew this would happen. Most of my math is self-taught, so there may be huge, unexpected holes in my knowledge...
  9. Oct 20, 2004 #8
    dont worry, i myself didnt know much about it (the teacher obviously skipped this issue in class).
    and anyway, if you dont know something there is always the web.

    now there real question is there any other way of prooving this question and if not does this approach is appropiate for any n degree equations when you have to use (a+b)^n+1 identity, if yes how do you prove it?
  10. Oct 20, 2004 #9
    Oh, yes, that is exactly what I mean, thanks a lot~ :smile:

    You also made me learned the sum and subtraction properties of series, I mean: [tex]\sum (n+m) = \sum n + \sum m [/tex]
    But, I still cannot get the exact same answer as yours. Mine is [tex]\frac {2n^3+3n^2+n-1}{6}[/tex]... dunno why, but there always is an extra -[tex]\frac{-1}{6}[/tex]

    I also tried the next question on my problem sheet, which is to prove [tex]\sum _ {r=1} ^n {r^3} = [\frac {n(n+1)}{2}]^2[/tex]. By using the same method, I found it is not too difficult to prove...

    En, not really, but there is a probability of 50% to predict if I am in Beijing or London a year, or 31 557 600 seconds :biggrin:
    Last edited: Oct 20, 2004
  11. Oct 20, 2004 #10
    Last edited: Oct 20, 2004
  12. Oct 20, 2004 #11
    Intuitively, I think that is appropriate (does not make much sense...)

    Anyway, big news, I found a simple mistake in my previous calculation and now the answers corresponds exactly.

    Thanks, Gokul~~~
  13. Oct 20, 2004 #12
    By the way, I want to ask that if [tex]\sum {nm} = (\sum {n})( \sum{m})[/tex] is also true?

    It would be great if it is true because the next question on my problem sheet is to sum [tex]\sum {(r+1)(r+3)}[/tex]...
    Last edited: Oct 20, 2004
  14. Oct 21, 2004 #13


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    No, it's not true.

    Consider the 2-term sum : a1b1 + a2b2. Clearly, this is different from (a1 + a2)(b1 + b2) .

    What you want to do is expand the product and sum up each of the terms in the expansion.
  15. Oct 21, 2004 #14
    Oh, thanks, you are right. It is not as tough as it looks... :redface:

    Well, here is the really tough one: [tex] \sum_ {n=0}^{n-1}(a+nd)r^n[/tex]

    I have tried to subtract [tex] r\sum_ {n=0}^{n-1} [/tex] from [tex]
    \sum_{n=0}^{n-1}[/tex]... and to subtract [tex] \frac {\sum_ {n=0}^{n-1}}{r}[/tex] from [tex] \sum_{n=0}^{n-1}[/tex] and see if something will comes out, however, nothing, comes out...
  16. Oct 21, 2004 #15


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    I hope you mean [itex] \sum_ {n=0}^{N-1}(a+nd)r^n[/itex]

    This is simply [itex]a\sum_ {n=0}^{N-1}r^n + d\sum_ {n=0}^{N-1}nr^n = aS1(N-1) + dS2(N-1)[/itex] where S1 is simply a geometric progression (GP) and S2 is an arithmetico-geometric progression (AGP).
  17. Oct 22, 2004 #16
    You should look into the book A=B by DOron Zeilberger et al. The methods exaplained are not that difficult and will provide you with everything you need to solve these Hypergeomtric identities. Basically it all boils down to convert the term in the sum as a recursion. Than you get a telescopi sum and the answer follows very naturally.

    The book is freely downloadable from the web.


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