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Math Help - Quadratic Expansion using Binomial Theorem

  1. #1
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    Quadratic Expansion using Binomial Theorem

    Hi, I am stuck with a question about Binomial Expansion using Binomial Theorem, here is the question:

    Find, in ascending powers of x, the first 4 terms in the expansion of (1 - 2x - 4x^2)^11

    Here is how I work out the question:
    [1 - 2x(1 + 2x)]^11
    = 1 + (11C1)[-2x(1+2x)] + (11C2)[-2x(1+2x)]^2 + (11C3)[-2x(1+2x)]^3 + .........
    = 1 + 11[-2x(1+2x)] + 55[-2x(1+2x)]^2 + 165[-2x(1+2x)]^3 + ............
    = 1 - 22x - 44x^2 + 55[(2x)^2 - 2(2x)(4x^2) + (4x^2)^2] + 165(-2x - 4x)^3 + ....
    = 1 - 22x - 44x^2 + 220x^2 - 880x^3 +880x^4
    + 165[(-2x)^3 + 3(-2x)^2(-4x^2) + 3(-2x)(-4x^2)^2 + (-4x^2)^2
    = 1 - 22x - 44x^2 + 220x^2 - 880x^3 +880x^4
    + 165(-8x^3 + 48x^4 + 96x^5 - 64x^6)
    = 1 - 22x + 176x^2 - 880x^3 + 880x^4 - 1320x^3 + ...............
    = 1 - 22x + 176x^2 - 2200x^3

    The bold part is the one that is wrong, I am not sure how I do it wrongly, can anyone guide me? thanks

    the solution is 1 - 22x + 176x^2 - 440x^3 + .......
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  2. #2
    Super Member ILikeSerena's Avatar
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    Re: Quadratic Expansion using Binomial Theorem

    Welcome to MHF, danny2012!

    Quote Originally Posted by danny2012 View Post
    = 1 + 11[-2x(1+2x)] + 55[-2x(1+2x)]^2 + 165[-2x(1+2x)]^3 + ............
    = 1 - 22x - 44x^2 + 55[(2x)^2 - 2(2x)(4x^2) + (4x^2)^2] + 165(-2x - 4x)^3 + ....[tex]
    You kept the minus sign in this step when you shouldn't have.

    [-2x(1+2x)]^2 = (2x)^2(1+4x+4x^2)
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  3. #3
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    Re: Quadratic Expansion using Binomial Theorem

    Hi ILikeSerena, thanks for your warm welcome

    You kept the minus sign in this step when you shouldn't have.
    [-2x(1+2x)]^2 = (2x)^2(1+4x+4x^2)
    I see, I forgot to include the bracket in -2x^2 which should be (-2x)^2 and the answer is 4x^2 lol thanks for the point out ILikeSerena
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