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Binomial Expansion

  1. Jan 20, 2007 #1
    1. The problem statement, all variables and given/known data

    Expand (1-6x)^4 (1+2x)^7 in ascending powers of x up to and including the terms in x^3


    2. Relevant equations

    (1-6x)^4 (1+2x)^7

    3. The attempt at a solution

    Firstly, I expand (1-6x)^4
    = (1)^4 + 4C1 (-6x) + 4C2 (-6x)^2 + 4C3 (-6x)^3 + ...
    = 1 + 4(-6) + 6 (36x^2) + 4(-216^3) + ...
    = 1 -24x + 216^2 - 864x^3 + ...


    Then after that, I expand (1+2x)^7
    = (1)^7 + 7C1 (2x) + 7C2 (2x)^2 + 7C3 (2x)^3 +...
    = 1 + 7 (2x) + 21 (4x^2) + 35 (8x^3) + ...
    = 1 + 14x + 84x^2 + 280x^3 + ...

    Finally, I expand (1-6x)^4 (1+2x)^7 together...
    = (1-24x+216x^2=864x^3+...) (1+14x+84x^2+280x^3+...)
    = 1+14x+84x^2+280x^3-24x-336x^2-2016x^3+3024x^3-864x^3+...
    = 1 -10x-252x^2+2240x^3+...

    My final answer is wrong.:frown: It should be 1-10x-36x^2+424x^3+...

    Can someone help me solve this? Thanks.:tongue:
     
  2. jcsd
  3. Jan 20, 2007 #2

    AlephZero

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    = (1-24x+216x^2=864x^3+...) (1+14x+84x^2+280x^3+...)

    That's OK

    = 1+14x+84x^2+280x^3 OK
    -24x-336x^2-2016x^3 OK
    +3024x^3-864x^3+... That's where you went wrong. There's a term missing, and probably you made another mistake adding up the terms because I can't see how you got 2240x^3 from your (wrong) numbers.
     
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