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[gr.11 math] help please.

  1. Jan 12, 2005 #1
    My teacher gave me a bonus question, and I really want to understand this and impress him lol. The current unit we are studying involves arithmatic/geometric sequences
    The question:

    We have to find the 10th term of:

    1/1 , 2/1 , -4/3 , 8/5 ...

    I want to understand this so please dont just give me the answer >.<

    What I know:
    Example of the hardest sequence I had to do so far:
    Determine a formula for the nth term of the sequence.
    1/2 , 3/4, 5/8, 7/16...[X]
    This is neither an arithmatic / geometric sequence (I'm not sure what it's called but im just going to call it a mixture of both)

    Numerator 1, 3, 5, 7... tn=1+(n-1)2 = 2n-1
    Denominator 2, 4, 8, 16... tn=2(2^n-1) = 2^n

    So, tn = 2n-1/2^n

    To find x = 2n-1/2^n = 2(5)-1/2^5 = 9/32

    a+(n-1)d (used to solve arithmatic sequence)
    a(r^n-1) (used to solve geometric sequence)
    ((a + tlast)2)/2
    tn = the number or the nth
    a = first number in sequence
    d = common difference
    r = common ratio

    EDIT: OMG DAMNIT!!! I am so sorry that I didn't post this in the right forum sry im newb and it's my first post and I didn't see that sticky argh! can a mod move this thread or??? Once again I apoligize for wrong forum :frown:
    Last edited: Jan 12, 2005
  2. jcsd
  3. Jan 12, 2005 #2
    So you are saying for the pattern [tex]\frac{1}{2} , \frac{3}{4} , \frac{5}{8} , \frac{7}{16}...[/tex] the equation for the [tex]n^{th}[/tex] term is [tex]\frac{2n-1}{2^n}[/tex]???

    (Just trying to make that clear.)

    The Bob (2004 ©)
  4. Jan 12, 2005 #3
    Yes that is what im saying.
  5. Jan 12, 2005 #4
    I can see that I was just making sure that we were talking about the same thing.

    Is this all you are given???

    The Bob (2004 ©)
  6. Jan 12, 2005 #5
    yeah that's all my teacher gave me
  7. Jan 12, 2005 #6
    I am not one of the best people to ask but I think that the denominator might be the best place to start.

    You have 1, 1, 3 and 5. The differences are 0, 2 and 2. Could the next difference be 4??? You see, I am think along the lines of 0 + 2 = 2 and so 2 + 2 = 4 and then 2 + 4 = 6. This might prove useful.

    The Bob (2004 ©)
  8. Jan 12, 2005 #7
    hmm that's very interesting you see our teacher hasn't taught us yet that we can use 0+2+2 as a common ratio or common difference, i thought it had to be one single value..hmm

    But if i follow what you said then i need a pattern for the top one which is 1,2,-4,8 which using your method could be 1x2x-2x-2...

    EDIT: i just can't figure out how to put this into a formula, because im almost positive there has to be a formula to do this...

    BTW BOB when you said 0+2+2 did you mean that i should keep repeating this as in:
    --and keep looping?
    So that the sequence would go like 1,1,3,5,5,7,9,9,11,13,13,15 ? I'm no expert but I'm not sure the method you suggested (Bob) is good. Although I'd be very happy if it were good but we can only find out if your method is good if you know a way to check our answer? But the only way we can check is if our pattern is correct, but how do we know if it's correct, ugh :cry:

    Unless i misinterpreted or you had a different idea in mind?
    Anybody else? please?
    Last edited: Jan 12, 2005
  9. Jan 12, 2005 #8
    You see that could be the pattern but with the numbers you have it is not easy to tell. There are lots of patterns that there would be. The demoniator (because it is the easiest) could be a 1 then a 1 then a 3 then a 5 and then another 5 and then a 7 and then a 9 (which is odd numbers but doubling each alternate one).

    Like I said I can only give you ideas and also if your teacher has not taught it and it is a bonus question then I think he is looking for some pattern that you have not studied yet.

    The Bob (2004 ©)
  10. Jan 12, 2005 #9
    That was an idea I was having but I was thinking more of the pattern 1, 1, 3, 5, 9, 15, 25, 41 etc...

    The Bob (2004 ©)
  11. Jan 12, 2005 #10
    heh, damn I wish math professors/teachers would see my post. I even wouldn't care if it was MY math teacher because at least he'd see how hard im trying, lol.
  12. Jan 12, 2005 #11


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    Here's what i give u:

    [tex] a_{n}=\frac{(-)^{n+1}2^{n}}{1+2(n-1)} [/tex]

    Check whether it fits all 4 terms you are given.If so,then u can use it to compute the value for "n=10".

  13. Jan 12, 2005 #12
    Alright well i tried your equation and it didn't work out for me but there is a high chance it's because i didn't do it correctly, can you correct me where i went wrong:

    an = (-)^n+1 x 2^n / 1+2(n-1) <--original.
    a1 = (-)^2 x 2^1 / 1+2(1-1)
    a1= -2/3(0)

    You see im not really sure what you meant by the (-) in the beginning of the equation, is that a -1? or ??
    I'm sry im a newb.
  14. Jan 12, 2005 #13
    yes, - means -1....
    you sure your sequence is 1/1 , 2/1 , -4/3 , 8/5 ??
    i can find a formulas to fit all these number but i don't think your teacher will ask you this question.
    if you want the a formulas fit this sequence and started at n=1, try this....
    [tex] a_{n}=\frac{(-1)^{n}2^{n-1}}{2n-3} [/tex]
    this fits the first 4 term.... but i doubt the first term in your sequence is negative in your problem....

    edit: dex's formulas started at n=0....
    Last edited: Jan 12, 2005
  15. Jan 12, 2005 #14
    Yes that is the sequence that my teacher gave me, thanks for your formula im trying it now and so far for the first term it is correct, im hoping it will work for all the other im checking now*

    But i still have a question, can you tell me how you got that formula?

    EDIT: DAMN! for n=2 it does not give the correct answer unless i plugged the numbers in incorrectly.

    a2 = (-1)^2 x 2^(2-1) / 2(2)-3
    a2 = -1(2)/1
    a2 = -2

    But it's supposed to equal 2! :confused:
    Last edited: Jan 12, 2005
  16. Jan 12, 2005 #15
    easy... do the numerator and the denominator separately.... if the sign alternate term by term.. you need (-1)^n, right? ....however, notice the signs is positive for the first 2 terms... also, the denominator is 1 for the first two terms.... what can we do is regard the first term as [tex] - 1/(-1)[/tex], then, the sequence become:
    [tex] -\frac{1}{-1} , \frac{2}{1} , -\frac{4}{3} , \frac{8}{5}...[/tex]
    you have an alternating signs sequence.... and do the numerator and denominator separately...
  17. Jan 12, 2005 #16
    [tex] (-1)^2 = [/tex] ?
    yes... you made an arithmatic mistake

    My formulas is fine.... you don't have to doubt it.... the question is.... do you understand how do I get this formulas.....
    Last edited: Jan 12, 2005
  18. Jan 12, 2005 #17


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    So it means that my formula works and it's okay... :tongue2: It means that i haven't waisted 10 minutes out my Earthly existence searching for it... :approve:


    PS.That [itex] (-) [/itex] instead of [itex] (-1) [/itex] is just routine...
  19. Jan 12, 2005 #18
    In my calculator (-1)^2= -1
    but if i do -1 x -1 then it gives me 1 which is correct.
    why does my calculator give me the wrong answer lol.

    But i still dont know how to get the formula....
  20. Jan 12, 2005 #19


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    Are u sure??When putting "-1" and then pressing the key/function "a squared" it gives u "-1"???Jesus,what kind of a beep-ed up computer is that?? :rofl:

    Inspiration and lots of work (practice/experience solving problems) are always the answers.And 'patience',let's not forget 'patience'.

  21. Jan 12, 2005 #20
    Remember to put it in brackets: (-1)^2. As -1^2 = (-)1^2 = -1
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