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The sum of series

  1. Feb 1, 2006 #1
    a series such as :-

    ∑ from n = 1 ,until m for n

    can be found with the law of the sum of arithmetical series:-

    where :-
    m= the number of terms
    a= the first term in the series
    d= the basic arithmetical

    in the pervious example :-

    a=1 , d=1 , m=m

    when we solve the last information in the law we find :-

    ∑ from n = 1 ,until m for n = m(m+1)/2

    for example :-
    1+2+3+4+..........+18+19+20 = 20(20+1)/2 = 210

    my question is :-

    what about the series :-

    ∑ from n = 1 ,until m for n^2 ???

    it isn't an artithmetical or even a geometrical series then :-

    what kind of series is it ?

    and how could they find that :-
    ∑ from n = 1 ,until m for n^2 = [m(m+1)(2m+1)]/6 ???

    for example :-

    (1^2)+(2^2)+(3^2)+(4^2)+(5^2)= 5(5+1)(2*5+1)/6 = 55

    and thank you
  2. jcsd
  3. Feb 1, 2006 #2


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    It's not an arithmetic series, obviously. In an arithmetic series, by definition, the difference between two consecutive terms must always be the same. But in 1, 22= 4, 32= 9, 42= 16, ..., the differences are 4-1 =3, 9- 4= 5, 16- 9= 7,...

    Notice, however, those differences are consecutive odd numbers: the "second differences" 5-3. 7- 5, ... are all 2. Obviously, for the sum of squares, the first difference is the squares themselves so the "third differences" are all 2. "Newton's divided difference formula" gives us
    [tex]\sum_{i=0}^n i^2= \frac{n(n+1)(2n+1)}{6}[/tex]

    Similar things can be done for higher powers but the results get progressively more difficult.
  4. Feb 15, 2006 #3
    Well sorry I was late to repost cause of exams

    well thank you mr.HallsofIvy for your answer, now it is more clear to me

    my reagrds
  5. Feb 19, 2006 #4
    If you want to put time into it, you can assume that the sum of the integers has a leading square factor, the sum of the squares has a leading cube term, the sum of third powers, etc. (This follows from HallsofIvy above.)

    We then attempt to find terms [tex]ax^3+bx^2+cx+d [/tex]=S(x)

    We know that S(0) = 0 and so d=0. S(1) = a+b+c=1, S(2)=8a+4b+2c=5, S(3) = 27a+9b+3c=14. So we can solve these equations by elimination of terms.
    Last edited: Feb 19, 2006
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