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Binomial Expansion

  1. Sep 29, 2014 #1
    1. The problem statement, all variables and given/known data

    Find the coefficient of x^3 in the binomial expansion of
    (2/x - 3x^4)^12

    2. Relevant equations


    3. The attempt at a solution

    Expanding this out would take too long and I cannot use a calculator to find the coefficient

    I know the formula for the expansion

    summation (12 choose k) a^k * b^12-k

    a = 2/x, b = -3x^4

    But how do I find k ?
     
  2. jcsd
  3. Sep 29, 2014 #2

    Orodruin

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    For which k does the term correspond to a x^3 term if you insert a and b into your expression?
     
  4. Sep 29, 2014 #3
    well if I ignore the coefficients, I get

    x^-k * x^(48-4k) = x^3
    48-5k = 3
    k = 9 ?

    so my coefficient would be 2^9 (-3)^3 ?
     
  5. Sep 29, 2014 #4

    Orodruin

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    Almost, you dropped the binomial coefficient which should also be there.
     
  6. Sep 29, 2014 #5

    HallsofIvy

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    You seem to have forgotten the "binomial coefficient", [tex]\begin{pmatrix}12 \\ 9 \end{pmatrix}[/tex].
     
  7. Sep 29, 2014 #6
    Ok, so with the binomial coefficient

    (12 choose 9) 2^9 (-3)^3 ?
     
  8. Sep 29, 2014 #7
    Okay guys, I have another question relating to the same topic

    Given (3x - 2/x^3)^40, Find coefficient x^10

    I'll skip the plugging into formula for a and b, but heres how I solve for k

    x^k * x^-3(40-k) = x^(-120+4k) = x^10
    -120+4k = 10
    k = 65/2

    Now in the memo, they have
    -120+4k = -20.........................How did they get -20 ?
    k = 100/4
     
  9. Sep 29, 2014 #8

    Orodruin

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    This is a very good question ... Just from looking at it for 5 seconds, I do not see the possibility of having a term x^10. Any term should be x^40 multiplied by some power of x^-4 which gives terms x^12 and x^8, but no term x^10.
     
  10. Sep 29, 2014 #9

    Ray Vickson

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    You wrote
    [tex] \left( \frac{2}{x} - 3 x^4 \right)^{12} [/tex]
    Is that what you meant, or did you want
    [tex] \left( \frac{2}{x - 3 x^4} \right)^{12}?[/tex]
    If the latter, use parentheses, like this: (2/(x - 3x^4))^12 or [2/(x - 3x^4)]^12.
     
  11. Sep 29, 2014 #10
    It's the first one
     
  12. Sep 29, 2014 #11
    So is the question wrong or something ?, I'm just not sure where the -20 came from
     
  13. Sep 30, 2014 #12
    Ok, the question was wrong, it was x^-20. All fine now, thanks guys :)
     
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