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    I don't get this question at all! Surely, every expansion in not independent of x?

    Also on the same paper, Q4c I've seen the mark scheme and It's not very helpful. Could somebody explain the concept/answer here?
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    Could you post the question please? Or give an exam board or something...
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    (Original post by naystar)
    Could you post the question please? Or give an exam board or something...
    http://www.mathsexams.ukteachers.com...2_2004_Jun.pdf
    Edexcel Maths P2 - Questions 4c and 3b
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    For 3b) the term independent of x is the one in the expansion where the power of x on top is equal to the power of x on the bottom.

    This would mean that x3a=xb
    3a=b
    At the same time you know that a+b=12
    so 4a=12, a=3
    b=9

    Then work out that term in the expansion.
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    question 4c anyone?
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    See attachment.
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    (Original post by Fermat)
    See attachment.
    cheers fermat any chance of Question 3b? It wasn't very well explained above :P
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    In the binomial expansion, the expansion is like this,

    (x + y)^n = x^n + nx^{n-1}y + n(n-1)x^{n-2}y^2/2{2!}+ ...

    and the rth (general) term is given by,

    C^n_{r-1}x^{n-(r-1)}y^{r-1}
    or,
    C^a_bx^{(a-b)}y^b

    where a = n, b = r-1

    We have \left(X^3 - \frac{1}{2X}\right)

    so, putting

    x = X^3
    y = -1/2X
    then rth term is
    C^a_b{X^3}^{(a-b)}{\left(\frac{-1}{2X}\right)}^b

    For the powers to cancel out,
    3(a-b) = b
    a = n = 12
    :. b = 9
    ======

    Coefft is C^a_b = C^{12}_9
    = 12!/9! = 12*11*10 = 1320

    You've also got to include (-1/2)^b from (-1/2X)^b which is (-1/2)^9 = -1/512.

    So coefft is 1320/(-512) = -165/64
 
 
 
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