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Putting into closed form

  1. Aug 5, 2008 #1
    Hi,

    I cant think / remeber how to write the following expresion in a closed form,

    the function is a summation of natural numbers between 1 and an upper limit "K", written as

    Sigma x with limits K and 1 effectivly, straightforward etc...
    what i want is the summation of all the "summations" between K and 1 so sigma (1,K) + sigma (1,(K-1)) + sigma (1,(K-2)) etc.. untill it reaches sigma (1,1) ie 1.

    its easy to visulise, take K as 5, the expression would be

    5+4+3+2+1
    +4+3+2+1
    +3+2+1
    +2+1
    +1

    which gives 35, i am wanting the answer in terms of K (im presuming its possible) or at least can be written much neater than an expansion, cheers (I thought it could possibly be written as a general sigma summation but prehaps with tending limits, but if so how do you represent the tending being discrete and not continous)

    cheers,
    tom
     
  2. jcsd
  3. Aug 5, 2008 #2

    HallsofIvy

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    It is fairly well known that
    [tex]\sum_{i=1}^n i= \frac{n(n+1)}{2}[/tex]
    so your "sum of sums" is
    [tex]\sum_{k=1}^n\left(\sum_{i=1}^k i\right)= \sum_{k=1}^n \frac{k(k+1)}{2}[/tex]

    It is also (reasonably) well known that
    [tex]\sum_{k=1}^n k^2= \frac{n(n+1)(2n+1)}{6}[/tex]

    putting those together,
    [tex]\sum_{k=1}^n \frac{k(k+1)}{2}= \frac{1}{2}\sum_{k=1}^n k^2+ \frac{1}{2}\sum_{k=1}^n k[/tex]
    [tex]= \frac{1}{2}\left(\frac{n(n+1)(2n+1)}{6}+ \frac{n(n+1)}{2}\right)= \frac{1}{4}n(n+1)\left(\frac{2n+1}{3}+ \frac{3}{3}\right)= \frac{n(n+1)(n+2)}{6}[/tex]

    You will note that for your example, with n= 5, this gives
    [tex]\frac{5(6)(7)}{6}= (5)(7)= 35[/tex]
     
    Last edited: Aug 5, 2008
  4. Aug 5, 2008 #3
    thats great thanks, i knew the general summations but tbh i wouldnt have the wit to make some of those connections, not currently at least...:P

    cheers
     
  5. Aug 5, 2008 #4

    HallsofIvy

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    That turns out be remarkably simple! I hadn't expected it to.
     
  6. Aug 5, 2008 #5

    HallsofIvy

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    Actually, I was surprised that the final formula was so simple!
     
  7. Aug 5, 2008 #6
    i recognise the final answer from somewhere else, o well thanks again
    :)
     
  8. Aug 6, 2008 #7
    can i just mention that that "proof" or process appears increadibly elegant, how long did it take for you to do, could i be so intrusive to ask about you level of expertise as i am taken back by again by the elegence of the solution

    cheers
     
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