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B Solve 3 7 12 18 25 series

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  1. Mar 2, 2017 #1
    Hi, im trying to solve this series generally:

    the series: 3 7 12 18 25.

    i tried using x(n) = 3 + 4n.

    But this doesnt work.. Please help.
     
  2. jcsd
  3. Mar 2, 2017 #2

    PeroK

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    I would add ##0## at the beginning and see whether that helps.
     
  4. Mar 2, 2017 #3

    mfb

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    Calculate how much the numbers increase in each step. That should lead to some pattern.
     
  5. Mar 2, 2017 #4

    OmCheeto

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    Did you try graphing it?
    When I graphed your series of numbers, I got a curve.
    That implied to me, that "4n", being a linear function, was not correct.
     
  6. Mar 2, 2017 #5

    Ssnow

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    You must find ##f(n)## such that ##a_{n+1}=f(n)+a_{n}## with ##a_{0}=3##, hint: think simple!
    Ssnow
     
  7. Mar 6, 2017 #6

    Svein

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    Most sequences can be resolve using difference (not differential) analysis.

    Code (Text):
         
     3
     7    4  
    12    5    1
    18    6    1
    25    7    1
     
     
  8. Mar 22, 2017 #7
    Just stumbled across this page. I'm doing independent math research and the series you are inquiring about is a smaller part of what I am doing. The formula you are looking for is F(n)= n(n+5) / 2 = {3,7,12,18,25,...}. It is similar to the formula for triangular numbers T(n) = n(n+1)/2 = {1,3,6,10,15,...}. Hope this helps!

    Regards,
    David
     
  9. Mar 24, 2017 #8
    Difference between two numbers is (2+n), so you can write x(n)= (2+n) + x(n-1) when n > 1.

    This is a recursive formula, that also can be written out as:
    x(n) = (2+n) + (2+n-1) + (2+n-2) + (2+n-3) + ... + x(1)

    Recognizing the pattern as:
    [itex]x(n) = x(1) + \sum_{i=2}^n(2+i) = x(1) - 3 + \sum_{i=1}^n(2+i) = \sum_{i=1}^n2+ \sum_{i=1}^ni = 2n+ n(n+1)/2 = n(n+5)/2 [/itex]

    So we find x(n) = n(n+5)/2.

    [itex]\sum_{i=1}^nn = n(n+1)/2[/itex] can be found from a table or WolframAlpha(or could also derivated without too much trouble) :)
     
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