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Finding Binomial Co-efficient from pronumerals

  1. Feb 10, 2012 #1
    1. The problem statement, all variables and given/known data

    I'm asked to find (a/b) in the simplest form if the co-efficient of x^8 is zero in the expansion of:

    (1 + x)(a - bx)^12


    2. Relevant equations

    Binomial expansion formula ... (a + b)^n = Sum of r --> n (r = 0) (nCr)(a^(n-r) * b^r



    3. The attempt at a solution

    I figured that x^8 could be achieved from two possible situations ...
    either 1 * the expansion of (a - bx)^12 or x * the expansion of (a - bx)^12

    I found the value of r at both these points by looking for the value of r that makes bx^r = x^8 and x*bx^r = x^8. This I found to be r = 7 and r = 8. Then I wrote as:

    (12C7) * (a)^5 * (-b)^7 + (12C7) * (a)^4 * (-b)^8 = 0

    Then I get stuck as I cannot seem to get values for a and b from this.

    Can anyone help me?
     
  2. jcsd
  3. Feb 10, 2012 #2
    Problem solved ...

    First of all, note my error in the last line

    (12C7) * (a)^5 * (-b)^7 + (12C7) * (a)^4 * (-b)^8 = 0

    Should be:

    (12C7) * (a)^5 * (-b)^7 + (12C8) * (a)^4 * (-b)^8 = 0

    The negative b on the (-b)^7 comes out the front as it is an odd power ... The other negative cancels out

    -(12C7) * (a)^5 * (b)^7 + (12C8) * (a)^4 * (b)^8 = 0

    throw the negative section over to the other side of the equals sign ...

    (12C8) * (a)^4 * (b)^8 = (12C7) * (a)^5 * (-b)^7

    then evaluate

    => (495)*(a^4)*(b^8) = (792)*(a^5)*(b^7)

    start cancelling

    => (495)*(b) = (792)*(a)

    divide to each side

    => 495/792 = a/b

    => a/b = 5/8

    which is correct according to the answers.
     
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