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Binomial expansion of term with x^2

  1. Feb 10, 2012 #1
    I have to determine the coefficient of an x term in an expansion such as this;
    Determine the coefficient of x^18 in the expansion of (1/14 x^2 -7)^16

    The general term in the binomial expansion is
    nCk a^k b^(n−k)
    I could let
    a = (1/14 x^2)
    b = -7
    n = 16
    k = 9?

    I have no real idea of how to go about finding this coefficient using the binomial theorem.

    Having expanded the expression to the 10th term I get

    8C9 (-7) (1/14 x^2)^9

    I'm using nCk = n! / (n-k)!k! but can't evaluate this as it is a negative.

    I'm assuming that the 8C9 bit is just the opposite of 6th term i.e. 12C5 = 792 (looking at Pascal's triangle this is on the opposite side), but I can get the x^18 bit (I'm assuming the (x^2)^9 can be x^18 here)

    Can someone check, please?
     
    Last edited: Feb 10, 2012
  2. jcsd
  3. Feb 10, 2012 #2

    Ray Vickson

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    Conventionally, nCk for integers 0 < n < k is regarded as zero. Why did you write 8C9?

    RGV
     
  4. Feb 11, 2012 #3

    HallsofIvy

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    If you let y= x^2, then you are looking for the coefficient of y^9 in ((1/14)y- 7)^16.
    That will be, of course, 16C9 (1/14)^9(-7)^7
     
  5. Feb 12, 2012 #4
    Of course you are right HallsofIvy.

    I was doing this

    (OK I put the 6th term when I meant 7th)
    I put 8C9 as I understand that when I expand the expression (1/14 x^2 - 7) longhand I get;

    -7^16 + (16C1*-7^15*(1/14 x^2)) + (15C2*-7^14*(1/14 x^2)) + . . . . + (11C6*-7^10*(1/14 x^2)) + . . . . . . . . + (8C9*-7^7*(1/14 x^2)) . . . .

    Where (11C6*-7^10*(1/14 x^2)) is the 7th term of the expansion
    & (8C9*-7^7*(1/14 x^2)) is the 10th term

    Pascals triangle, being symmetrical, should reflect the coefficients around the8th & 9th terms. So I'm assuming that the coefficient of the 7th term should be the same as the 10th term.

    When I try to calculate 8C9 (which I did longhand, shown above)
    I use form nCk = n! / (n-k)!*k!
    If I let n=8
    & k=9
    I get 8!/(-1)!*9!

    I work this out to be;
    40320 / -1 * 362880 = -0.1

    I know this isn't right so why did I do this wrong!
     
  6. Feb 12, 2012 #5

    HallsofIvy

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    [itex]_8C_9[/itex] would be the coefficient of [itex]a^9[/itex] in [itex](a+ b)^8[/itex] and there is no such term! Do you mean [itex]_9C_8[/itex]?
     
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