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Sequences Series geometric series or an arithmetic series?

  1. Apr 8, 2007 #1
    This is the sequence: 1, 2, 5, 14, 41, 122

    1. Is this a geometric series or an arithmetic series?
    2. I know the formula is a sub n=[3^(n-1)+1]/2, but how do you get that from a sub n=a sub 1 * r^(n-1), which is the geometric formula for series.
    Last edited: Apr 8, 2007
  2. jcsd
  3. Apr 8, 2007 #2


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    1. Neither.

    2. I assume your sequence should be 1,2,5,14,41,122

    You don't get it from a geometric series, since it isn't a geometric sequence.
  4. Apr 8, 2007 #3


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    It is neither - geometric series means constant ratio between terms, arithmetic means constant difference.
  5. Apr 8, 2007 #4
    So how would you get it then, with what formula or method?
  6. Apr 8, 2007 #5


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    Well, if you are clever, you will see that if n>=2, then we have:


    Assuming this is the pattern for all the next numbers, you may derive that explicit formula.
  7. Apr 8, 2007 #6
    So there is no definite way of determining that formula? I see that to get from a term to another you add first 1, then 3, then 9, then 27, then 81 which are multiples of three. Does this have anything to do with determining the formula?

  8. Apr 8, 2007 #7


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    A finite sequence can be extended in infinitely many ways, i.e, there exists an infinity of patterns to choose from.
  9. Apr 8, 2007 #8
    A pattern that fits this is: [tex]F(N)=\frac{3^N+1}{2}, N=0,1,2..[/tex]
  10. Apr 9, 2007 #9
    I know that that is the pattern, but I was just wondering how to figure that out with a formula or something.
  11. Apr 9, 2007 #10

    2) you can convert it to geometric series >>

    t=1+2+5+14+41+122..... Tn ------(1)
    t= 1+2+5+14+41+122....Tn-Tn-1+Tn ------(2)

    now eq 1- 2 and you'll get geometric series .
  12. Apr 9, 2007 #11
    Can you please explain that more?
  13. Apr 9, 2007 #12

    matt grime

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    Post 5 gives you a difference equation to solve. You also have the solution as a closed formula, to the difference equation, so it is a simple induction argument to verify it.
  14. Apr 9, 2007 #13
    And yet, in the context of a math education, only one of these is considered to have a "formula". Although we with our modern sensibilities abhor this notion of formula, Euler would concur.
  15. Apr 10, 2007 #14

    Tn=1+2+5+14+41+122..... Tn ------(1)
    Tn=0+1+2+5+14+41+122....Tn-1+Tn ------(2)

    -------------------------------------------------------------- Eq 1- 2

    0 = 1+1+3+3^2+3^3+3^4+................ (Tn-Tn-1) - Tn

    now transfer that Tn to that side (where zero is) & other side will have (n-1) terms and if you'll not include 1 of (first one) of series then terms will be (n-2) .that fact is that when you get such type of serieses you have to see for diffrences of series .

    Last edited: Apr 10, 2007
  16. Apr 12, 2007 #15

    Usualy, assuming that we deal with a homogeneous linear sequence, the recurrence relation which we have to seek is that having the smallest degree, in this case a[k]-4a[k-1]+3a[k-2]=0 which gives immediately the closed form from the OP (k=3,4...).

    [The characteristic equation is r^2-4*r+3=0 ---> r1=1; r2=3

    Therefore we must seek a solution of the form a[k]=A*(1)^k+B*[(3)^k] (1); A,B = constants

    we have a[k=3]=5 and a[k=4]=14 ---> replacing k in (1) with 3 and 4 results a system of equations from which A=1/2 and B=1/6.]
    Last edited: Apr 12, 2007
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