nth term of a sequence....

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  1. flown_muse's Avatar
    1 18-03-2012  18:46
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    nth term of a sequence....
    [Edit]
    Last edited by flown_muse; 18-03-2012 at 21:13.
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  2. Cerdog's Avatar
    2 18-03-2012  18:53
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    Re: nth term of a sequence....
    Set n = 2k, and consider that -8 = (-2)^3. You should be able to make some progress from there (unless I've made a misunderstanding somewhere).
  3. f1mad's Avatar
    3 18-03-2012  18:55
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    Re: nth term of a sequence....
    Yeah that sounds right.
  4. flown_muse's Avatar
    4 18-03-2012  19:10
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    Re: nth term of a sequence....
    [Edit]
    Last edited by flown_muse; 18-03-2012 at 21:14.
  5. Cerdog's Avatar
    5 18-03-2012  19:14
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    Re: nth term of a sequence....
    (Original post by flown_muse)
    Okay, now I have:

    a_2_k = \frac{(2k-1)(-2)^3^(^2^k^+^1^)}{2^(^2^k^-^1^)}

    and now I have an issue, because I can't find a value of k that can make both the top and the bottom equal to a power of 4.

    I'm clearly doing something wrong
    Write it as a_2_k = (2k-1)\frac{(-2)^3^(^2^k^+^1^)}{2^(^2^k^-^1^)}, and remember that -2 = -1 \times 2, so you can expand the powers to get something easier.
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  6. f1mad's Avatar
    6 18-03-2012  19:16
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    Re: nth term of a sequence....
    (Original post by flown_muse)
    Okay, now I have:

    a_2_k = \frac{(2k-1)(-2)^3^(^2^k^+^1^)}{2^(^2^k^-^1^)}

    and now I have an issue, because I can't find a value of k that can make both the top and the bottom equal to a power of 4.

    I'm clearly doing something wrong
    Expand the numerator and simplify.
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    1
  7. flown_muse's Avatar
    7 18-03-2012  19:18
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    Re: nth term of a sequence....
    [Edit]
    Last edited by flown_muse; 18-03-2012 at 21:15.
  8. f1mad's Avatar
    8 18-03-2012  19:39
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    Re: nth term of a sequence....
    (Original post by flown_muse)
    a_2_k = (2k-1)\frac{(-2)^3^(^2^k^+^1^)}{2^(^2^k^-^1^)}
    a_2_k = (2k-1)\frac{(-1)(2)^(^6^k^+^3^)}{2^(^2^k^-^1^)}
    a_2_k = (2k-1).(-1)(2)^(^6^k^+^3^)^-^(^2^k^-^1^)
    a_2_k = (2k-1).(-1)(2)^(^4^k^-^2^)

    etc.?
    What's 3--1?

    Apart from that yeah that's right. You can get rid of the 1 btw.
    Last edited by f1mad; 18-03-2012 at 19:43.
  9. flown_muse's Avatar
    9 18-03-2012  19:43
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    Re: nth term of a sequence....
    (Original post by f1mad)
    Yeah that's right. You can get rid of the 1 btw.
    So, I have ended up with:

    a_2_k = (2k-1).(-1)(2)^(^4^k^+^4^)
    a_2_k = (1-2k).2^4^k^+^4

    Therefore, 4k+4=4, 4k=0
  10. f1mad's Avatar
    10 18-03-2012  19:50
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    Re: nth term of a sequence....
    (Original post by flown_muse)
    So, I have ended up with:

    a_2_k = (2k-1).(-1)(2)^(^4^k^+^4^)
    a_2_k = (1-2k).2^4^k^+^4

    Therefore, 4k+4=4, 4k=0
    Hang on, where did the last line come from?
  11. flown_muse's Avatar
    11 18-03-2012  19:51
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    Re: nth term of a sequence....
    (Original post by f1mad)
    Hang on, where did the last line come from?
    Because the question wants me to find An in terms of powers of 4, so I'm setting the power equal to 4.

    (I think?)
  12. flown_muse's Avatar
    12 18-03-2012  19:59
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    Re: nth term of a sequence....
    Oh. I never thought of that. I think you are right, because otherwise I end up with n=0 and the answer is just a number with no powers. Argh! Thanks
  13. f1mad's Avatar
    13 18-03-2012  20:01
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    Re: nth term of a sequence....
    (Original post by flown_muse)
    Oh. I never thought of that. I think you are right, because otherwise I end up with n=0 and the answer is just a number with no powers. Argh! Thanks
    Yeah you need to re-write it . In terms of 4^ something, since this would then imply powers of 4.
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