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Summation of series

  1. Jul 14, 2011 #1
    Find the sum of n terms of the series:

    7/(1.2.3) - 17/(2.3.4) + 31/(3.4.5) - 49/(4.5.6) + 71/(5.6.7) - ...


    I know how problems like the following are solved :
    1. 1/(1.2.3) + 2/(2.3.4) + 3/(3.4.5) + ...
    2. 3/(1.2.4) + 4/(2.3.5) + 5/(3.4.6) + ...


    What will be the general term of the required series? How do I proceed exactly? I am not understanding how to start solving the problem.
     
    Last edited: Jul 14, 2011
  2. jcsd
  3. Jul 14, 2011 #2
    Series can definitely be tricky.

    Things to look at first - with a series such you need to look at distinct patterns in the numbers given to you. The first thing that jumps out at my eye is that the sigh (+ or -) alternates each term. What way do series that do that are generally represented?

    Next I would tackle this part:

    [tex]\frac{1}{1*2*3} + \frac{1}{2*3*4} + \frac{1}{3*4*5} [/tex] and see if you can't come up with a separate summation representation for that. At this point, you almost have all of it done.

    The next part is to look at the integer portion at the top, so the [7, 17, 31, 49] part of the terms. See any relationships between them?

    Hopefully that helps you get started.
     
  4. Jul 14, 2011 #3
    If you consider the difference between the difference of the consecutive terms, it comes out to be constant.
    7 17 31 49
    10 14 18
    4 4

    Now what?
     
  5. Jul 14, 2011 #4
    Perfect. Now how would you incorporate that with getting those particular initial terms in a sum? How could you then make the sign change each time?
     
  6. Jul 17, 2011 #5
    We need a (-1)^n term in the general term for the sign change, that is okay..
    Now, I know 2 things :
    1: 17 = 7+10
    2: 31 = 17 + 14
    3: 14 = 10 + 4
    Let me put the above two equations in variable form.

    Consider this -

    7 = x1
    17 = x2
    31 = x3


    10 = y1
    14 = y2

    and

    4 = z1

    Basically,

    1 => x2 = x1 + y1

    2=> x3 = x2 + y2

    3=> y2 = y1 + z1

    Therefore, we have -

    x3 = x1 + 2y1 + z1
     
  7. Jul 18, 2011 #6
    Don't look at the difference between the difference of the terms. If you just consider the difference of the terms, they form an arithmetic sequence: 10, 14, 18, 22, ...
    Now take a look at this:

    a2 = a1 + 10
    a3 = a2 + 14 = a1 + 10 + 14
    a4 = a3 + 18 = a1 + 10 + 14 + 18

    It just becomes a1 plus an arithmetic series.

    Since a1 = 7, try plugging everything into the formula Sn = n(a1 + an)/2. You'll have to play around a bit with the formula you get, but then that'll take care of the numbers in the numerator.
     
  8. Jul 19, 2011 #7
    To me, this looks like -
    Sn = a1 + 5*2 + 7*2 + ... (till n terms)
    and an = a1 + 2*(2n+1) ; n is a natural number that lies between 2 and n.

    So, Sn = [n(2a1 + 4n + 2)] / 2

    (Since Sn = n(a1 + an) / 2 )

    Therefore, Sn = n[16 + 4n] /2
    = 8n + 2n^2 ( n going from 2 to n)

    Is that correct?
     
    Last edited: Jul 19, 2011
  9. Jul 19, 2011 #8

    vela

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    If you plug in values for n, does your formula reproduce the sequence you're seeking?
     
  10. Jul 19, 2011 #9
  11. Jul 19, 2011 #10

    vela

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    Then it's correct.
     
  12. Jul 19, 2011 #11
    Oh wait, I got the summation only for the numerator part.
     
  13. Jul 19, 2011 #12
    Well, see if you can take a stab at it.

    The nature of the numbers should make you think factorials. Using that intuitive leap, try to logic out a way to - as vela suggested - plug in n and receive the correct values.
     
  14. Jul 19, 2011 #13
    Don't forget the a1 in the summation formula to get the numbers for the numerators. If bn is the sequence of numbers for the numerators, bn is basically going to be Sn once you've plugged in a1. Then you just have the denominators to finish, and that's easy. :smile:
     
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