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Math Help - 20th term in sequence

  1. #1
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    20th term in sequence

    Find the next and 20th term in the sequence.

    1, -3, 2, -5, 4, -7, 8, -9...

    This one seems very hard to me. I tried the difference table method, which gives no pattern that I can see.

    1 -3 2 -5 4 -7 8 -9
    -4 5 -7 9 -11 15 -17
    9 -12 16 -20 26 -32
    -21 +28 -36 46 58

    Only thing I can think of so far is the alternate +ve and -ve values could be solved with (-1)^n, so it gives a negative value for odd powers.

    This problem does not seem to be a simple arithmetic or geometric sequence. Can anyone guide towards what area of mathematics this problem is from, and is there a way to think of similar sequence problems?

    Thanks.
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  2. #2
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    e^(i*pi)'s Avatar
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    Your odd terms in the sequence are doubling from the previous one

    Your even terms form an arithmetic sequence with a=-3 and d=-2.

    Can you find the next (odd) term and the 20th (even) term given the above?

    edit: the question does not ask for the nth term so don't worry about it!
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  3. #3
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    Quote Originally Posted by mathguy80 View Post
    Find the next and 20th term in the sequence.
    1, -3, 2, -5, 4, -7, 8, -9...
    Try a_n  = \left\{ {\begin{array}{rl}<br />
   {2^{\frac{{n - 1}}<br />
{2}} ,} & {\text{if n is odd}}  \\<br />
   { - n - 1,} & {\text{if n is even}}  \\ \end{array} } \right.
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  4. #4
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    Didn't think of it that way at all! So its both an arithmetic sequence for the common difference in even terms and geometric for the odd terms?

    So for odd terms its 2^(n-1) and for even terms it's -2n-1. So depending on if the odd/even term is required use corresponding function.

    Is this correct, Having 2 different functions for the same sequence?
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  5. #5
    Super Member TheChaz's Avatar
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    Techincally, it's just one function, where even and odd terms are defined separately. This is certainly still a (well-defined) function.

    Suffice it to say "yes", this is correct!
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  6. #6
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    Quote Originally Posted by mathguy80 View Post
    So for odd terms its 2^(n-1) and for even terms it's -2n-1. So depending on if the odd/even term is required use corresponding function.
    That does not work.
    When n=3 it is odd.
    BUT 2^{3-1}=4, and that is not the third term.

    It is quite common to use two or more expressions to define sequences.
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  7. #7
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    Hello, mathguy80!

    \text{Find the next and 20th term in the sequence: }\;1, \text{-}3, 2, \text{-}5, 4, \text{-}7, 8, \text{-}9 \hdots

    e^{1\pi} is correct.

    There are two sequences which have been riffle-shuffled together.

    . \begin{array}{c||c|c|c|c|c|c|c|c}<br />
n & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline \hline<br />
\text{odd} & 1 && 2 && 4 && 8 & \\ \hline<br />
\text{even} && \text{-}3 && \text{-}5 && \text{-}7 && \text{-}9 \\ \hline \end{array}


    And Plato's piecewise function is perfect.


    And now for my two cents . . .
    The function can be written: . a_n \;=\;\dfrac{1\!-\!(\text{-}1)^n}{2}\cdot 2^{\frac{n-1}{2}} - \dfrac{1\!+\!(\text{-}1)^n}{2}\cdot(n+1)

    This result can be used to please/surprise/impress/terrify your teacher.

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  8. #8
    Super Member TheChaz's Avatar
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    Quote Originally Posted by Soroban View Post
    ...
    This result can be used to please/surprise/impress/terrify your teacher.
    I am pleased and impressed!
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  9. #9
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    Wow! That is a beautiful function. I see what you mean about the error in mine. Thanks for your help, you have shown me a different way of looking at sequences today.
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