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Math Help - binomial coefficient.

  1. #1
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    binomial coefficient.

    So I'm trying to find x^13 and x^14 in (4x^3-(1/x))^24

    I've done:
    (24)
    (k) (4x^3)^K (-1/x)^(24-K)

    (24)
    (k) 4^Kx^3K (-1)^(24-k) x^(K-24)

    (24)
    (k) 4^K (-1)^(24-K) x^(3K+K-24)

    (24)
    (K) 4^K (-1)^(24-K) x^(4K-24)

    and so for the 13th coefficient, 13 = 4K-24
    which is 4K=37, but there's no integer answer for K? I'm sure this is wrong, please help!

    PS at the start of each line the 24 and the K are together (ie in two brackets, I'm sure you'll get it)
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  2. #2
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    Quote Originally Posted by brumby_3 View Post
    which is 4K=37, but there's no integer answer for K? I'm sure this is wrong,
    from your working this answer is fine, if k \notin \mathbb{Z } it means there is no x^{13} term
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  3. #3
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    Thanks!
    For x^14:

    14 = 4K-24
    4K = 38
    and that doesn't have an integer answer either. Is this right? So both x^13 and x^14 do not appear in the expansion, right? (From my lecture notes, the lecturer said there is no coefficient if it's not an integer answer.)
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  4. #4
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    Looks like your lecturer and I agree. Best way to know for sure is to expand it out. If you have a good calculator or a computer package like Maple, it won't take long at all.
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