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Math Help - Finding the value of a term

  1. #1
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    Finding the value of a term

    I am stuck on a problem here that I have been looking at since last week. If someone could guide me in the right direction I would appreciate it before my head explodes!

    SEE ATTACHED DOCUMENT: I did not realize when I copied and pasted the original equation it changed it to the same size font. It was an intelligible equation when I typed it.

    I am utterly lost. Where would I even begin in order to solve?
    Thanks for any help!
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    Last edited by mathfanatic; September 16th 2008 at 12:39 PM.
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  2. #2
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    Quote Originally Posted by mathfanatic View Post
    I am stuck on a problem here that I have been looking at since last week. If someone could guide me in the right direction I would appreciate it before my head explodes!

    a1=3, ak+1=ak-2
    What is the value of the 4th term?
    I am utterly lost. Where would I even begin in order to solve?
    Thanks for any help!
    a_{k+1}=a_{k-2}

    so if k=3, we have: a_4=a_1=3

    RonL
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  3. #3
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    Quote Originally Posted by CaptainBlack View Post
    a_{k+1}=a_{k-2}

    so if k=3, we have: a_4=a_1=3

    RonL

    Did you get k=3 from the a1=3?

    I am sorry, I am so lost with this!
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  4. #4
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    You need to write this problem more clearly to make it intelligible. It's obviously about a sequence of terms a_1,\,a_2,\,a_3,\,a_4,\ldots, and you're given two equations to enable you to calculate these terms. The first equation presumably says a_1=3. In other words, the first term in the sequence is 3 (and the 1 in that equation is a subscript). I'm guessing that the other equation is meant to tell you how to find a_{k+1} if you know a_k. In other words, the k+1 on the left-hand side is a subscript; but on the right-hand side, only the k is a subscript, not the -2. Am I right so far?

    If I am, then the second equation probably says either a_{k+1} = a_k-2 or a_{k+1} = a_k^{-2}. If it's the first of those, then it is saying that you subtract 2 from each term of the sequence to get the next term. If the -2 is meant to be a power of a_k, then it's saying that you must raise each term of the sequence to the power -2 to get the next term.

    Whichever of those two versions is correct (or maybe it's something else altogether?), you apply that rule to the first term a_1 (namely the number 3) in order to get the second term a_2, and then repeat the same process to get a_3 and finally a_4.

    Edit. Another possible interpretation is the one that CaptainBlack is suggesting, namely a_{k+1} = a_{k-2}, with the whole of the k-2 as a subscript. In that case, each term is equal to the one that came three terms before it (because the difference between k+1 and k-2 is 3). In that case, a_4=a_1 as he says.
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  5. #5
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    Hello, mathfanatic!

    It's impossible to read what you typed,
    . . but I'll take an educated guess . . .


    a_1\:=\:3,\quad a_{k+1} \:=\:a_k-2 . . . I hope this is right!

    What is the value of the 4th term?

    It says (I think): . a_{k+1} \;=\;a_k - 2

    That is, each term is two less than the preceding term.
    [The sequence "goes down by 2's."]


    Then: . \boxed{\begin{array}{ccc}a_1 &=& 3 \\ a_2 &=& 1 \\ a_3 &=& \text{-}1 \\ a_4 &=& \text{-}3 \\ a_5&=& \text{-}5 \\ \vdots &&\vdots\end{array}}

    Got it?

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  6. #6
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    Thank you!!

    Thanks-yes I didn't realize when I pasted it from Word that it changed the font-I attached it as a document later.

    But this is a PERFECT explanation!! You have been absolutely a wonderful help-I am so happy!!

    Thank you thank you thank you!!!!! You made my day!!


    Quote Originally Posted by Soroban View Post
    Hello, mathfanatic!

    It's impossible to read what you typed,
    . . but I'll take an educated guess . . .


    It says (I think): . a_{k+1} \;=\;a_k - 2

    That is, each term is two less than the preceding term.
    [The sequence "goes down by 2's."]


    Then: . \boxed{\begin{array}{ccc}a_1 &=& 3 \\ a_2 &=& 1 \\ a_3 &=& \text{-}1 \\ a_4 &=& \text{-}3 \\ a_5&=& \text{-}5 \\ \vdots &&\vdots\end{array}}

    Got it?
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