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Math Help - [SOLVED] Binomial Theorem question...

  1. #1
    Member looi76's Avatar
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    [SOLVED] Binomial Theorem question...

    Question:
    Expand (1-3x)^10 up to and including the term x. Deduce the coefficient of x in the expansion of (1+3x)^2(1-3x)^10.

    Attempt:
    (1-3x)^10 = 10C0*1^10 + 10C1*1^9*(-3x) + 10C2*1^8*(-3x)^2 +.....
    (1-3x)^10 = 1*1 + 10*1*(-3x) + 45*1*(-3x)^2 +.....
    (1-3x)^10 = 1 - 30x + 405x +.....

    (1+3x)^2 = 2C0*1^2 + 2C1*1*(3x) + 2C2*(3x)
    (1+3x)^2 = 1*1 + 2*1*3x + 1*9x
    (1+3x)^2 = 1 + 6x + 9x

    = (1 + 6x + 9x)(1 - 30x + 405x +.....)
    Coefficient of x = 405 + 180 + 9
    Coefficient of x = 594

    Answer in textbook is 234, what is wrong?
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  2. #2
    GAMMA Mathematics
    colby2152's Avatar
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    Quote Originally Posted by looi76 View Post
    = (1 + 6x + 9x)(1 - 30x + 405x +.....)
    Coefficient of x = 405 + 180 + 9
    Coefficient of x = 594

    Answer in textbook is 234, what is wrong?
    I didn't check the beginning, but I know you went wrong here...

    (1 + 6x + 9x^2)(1 - 30x + 405x^2) =
    = 1 - 30x + 405x^2 - 30x - 180x^2 + 6(405)x^3 + 9x^2 - 270x^3 + 9(405)x^4

    (405-180+9)x^2 = 234x^2
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