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Technique in Sequence.?

  1. Mar 8, 2009 #1
    Can anyone help me how to get the formula fast?

    Example:
    1) 2, 4, 8, 16 n=1
    Answer: 2^n

    Problem:
    1) -2, 2, 22, 122... n=0
    2) 8, 19, 41, 85, 173, 349.... n=1
    3) -1/3, 1/2, -3/4.... n=1
     
    Last edited: Mar 8, 2009
  2. jcsd
  3. Mar 8, 2009 #2

    Mentallic

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    I am really inexperienced at sequences, especially when it comes to finding formulas; so please excuse my help as it could be a very unconventional way to approach the problems.

    1) I don't know

    2) I noticed that the difference between each term started with 11, and doubled each time so I assumed the formula would be in the form of [tex]y=a.2^{x+b}+c[/tex] and after some guessing, I let a=11 because the first difference was 11... edit: broke a rule.
    3) The sequence is being multiplied by [tex]\frac{-3}{2}[/tex] since [tex]\frac{S_2}{S_1}=\frac{S_3}{S_2}=\frac{-3}{2}[/tex] but I'm unsure where to go from there :blushing:
     
    Last edited: Mar 8, 2009
  4. Mar 8, 2009 #3

    tiny-tim

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    Hi sarah22! :smile:

    In problems like this, the methods that usually work are to write the numbers on one line, and either the differences or the ratios between successive numbers on their next line …

    then do it again and again until you get a line that's easy (like 2 2 2 2… or 1 3 5 7 …) …

    in at least one of your three problems, you'll have to use both differences and ratios (on different lines, of course) :wink:
     
  5. Mar 8, 2009 #4

    Mentallic

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    edit: broke a rule.
     
    Last edited: Mar 8, 2009
  6. Mar 8, 2009 #5

    tiny-tim

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    Mentallic, please do not give complete answers!!
     
  7. Mar 8, 2009 #6
    If the formula is a polynomial, try the Lagrange interpolating polynomial.
     
  8. Mar 8, 2009 #7
    Ok I got number 2 and some in 3. I'm sorry I didn't put where the n starts. I'll edit my first post.

    2) (11)(2^n) -3
    3) (-1/3)(-3/2)^n if n = 0. but it starts with 1 so how can I do it?

    I still don't get the other number. OMG. I'm doom.

    @Mentallic
    Ok I got some in number 3.

    @tiny-tim
    Yeah I'm doing that but still I don't get number 1 and 3.

    @bpet
    Can you teach how is it done?

    Thanks everyone for helping me. :wink:
     
    Last edited: Mar 8, 2009
  9. Mar 8, 2009 #8

    tiny-tim

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    Hi sarah22! :wink:

    Try 1) … write out the differences and you get … ? :smile:
     
  10. Mar 8, 2009 #9
    Ok here is it. I still don't get it. Oh no! I don't want to fail.

    0 -(2) ?(n) n=0
    4 -(2) 4(n) n=1
    24 -(2) 12(n) n=2
    144-(2) 48(n) n=3
     
  11. Mar 8, 2009 #10

    tiny-tim

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    eugh! :yuck: you're adding! why are you adding? :cry:

    take the differences :smile:
     
  12. Mar 9, 2009 #11

    Mentallic

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    You want to make the first term to the power of 0, [tex](-1/3)(-3/2)^0[/tex]

    then the next term to the power of 1, [tex](-1/3)(-3/2)^1[/tex]

    but for the first term n=1 and for the second, n=2.
    Fiddle with the exponent a bit, I'm sure you'll get it :wink:

    For #3, when you take the differences between the terms, notice the rate at which the numbers are rising. Is it by a multiple, a ratio, an exponential..?
     
  13. Mar 9, 2009 #12
    Ok, I got a technique from my friend. I'll just subtract the first term from the next then see how is it made.

    -2, 2, 22, 122... n=0

    2-2, 22-2, 122-22
    0, 20, 100


    So on each term it is increase 5 times the previous.

    (5^n)
    1, 5, 25, 125

    so to make the formula correct. You need to minus it by 3. that's it.

    My professor didn't teach most of this technique. I really hate her!

    Yeah, I'm a bit crazy the time I posted it. lol. You just need to subtract n in 1.

    (-1/3)(-3/2)^n-1

    I dunno. Is it exponential because I raise it by n?
     
  14. Mar 9, 2009 #13

    tiny-tim

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    did you do this yourself, or just copy it? either way you got it wrong :redface:

    it's 2 - (-2) = 4, not 0 …

    and how did you get
    from 0, 20, 100 ?
    uhhh? where does the 3 come from? and what do you mean anyway?
    tush, sarah22 … that's no way to talk about your nice teacher! :rolleyes:

    anyway, show us the formula, just to make sure you've got it now. :smile:
     
  15. Mar 10, 2009 #14

    Mentallic

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    So even though you got the difference between the first and second term wrong, you found the rate of increase correctly. Handy :smile:


    that's it? nearly. Test your formula [tex]S_n=5^n-3[/tex] and you'll see what you're missing.



    Yes :smile:
     
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