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Geometric Series question

  1. Jul 2, 2010 #1
    This looks almost like a geometric series;

    1, 2, 5, 14, 41, 122, 365, ...

    but each term is one less than three times the preceeding one. So is this a sequence or a series? What is a formula for the value of the nth term in terms of n?
     
  2. jcsd
  3. Jul 2, 2010 #2

    disregardthat

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    So basically you have

    (1) [tex] \ \ \a_{n+1} = 3a_n-1 \ \ \[/tex]

    and [tex]a_1 = 1[/tex].


    I suggest you try to expand [tex]a_n[/tex] using (1) given above, hypothesise a reasonable formula for [tex]a_n[/tex] and prove your suggestion by induction.

    A trick however: Set [tex]b_n = a_n- \frac{1}{2}[/tex]. In that case the formula (1) is reduced to [tex]b_{n+1} + \frac{1}{2} = 3(b_n + \frac{1}{2})-1 \Leftrightarrow b_{n+1}=3b_n[/tex]. This you can surely solve easily.
     
  4. Jul 3, 2010 #3
    You should write out what a(n) is in terms of a(1). It's quite easy. On part will be a geometric series.
     
  5. Jul 3, 2010 #4

    HallsofIvy

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    Do you not understand the difference between a sequence and a series? A series is a sum of numbers. It has nothing to do with being "geometric" or not. '1, 3, 9, 27, ... is a goemetric sequence. 1+ 3+ 9+ ... is a geometric series.

    What you give is a sequence because there is no sum. 1+ 2+ 5+ 14+ ... would be a series.
     
  6. Jul 3, 2010 #5
    ok, this looks like it's it,

    a ( n ) = [1 + 3^( n - 1 )] / 2 = 1, 2, 5, 14, 41, 122, 365, ...
     
  7. Jul 3, 2010 #6
    Well a series is a sequence if we take a series to be the limit of its partial sums, which in many scenarios is the case. But yeah I consider even basic questions like "is this a sequence or a series" to be unimportant. Basic logic indicates that it is a list of numbers, and not a sum, so who cares what it's called if you need to find the n-th term?
     
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