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nth term of a sequence....

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    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).
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    Yeah that sounds right.
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    (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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    (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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    (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.
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    (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
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    (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?
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    (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?)
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    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
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    (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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