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    4 part question

    1(a)Find first 3 terms in ascending powers of x of binomial expansion (2-9x)^4
    1st term: 16
    2nd term: -288x
    3rd term:1944x^2

    (b) f(x)=(1+kx)(2-9x)^4 where k is constant
    the expansion in ascending powers of x of f(x) up to and including the term x^2 is
    A-232x+Bx^2 where A and B are constants

    (1+kx)(16k-288x+1944x^2)
    16k^2x-288kx^2+1944x^3k

    this is my working out but it does not seem to lead anywhere any help much appreciated for part b and c and d.

    (c)Find value of K

    (d)Hence value for B
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    (Original post by Anonymous1502)
    (1+kx)(16k-288x+1944x^2)
    16k^2x-288kx^2+1944x^3k
    Are you sure that you should have a k in the second bracket before you expand?
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    (Original post by Pangol)
    Are you sure that you should have a k in the second bracket before you expand?
    Not sure.
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    (Original post by Anonymous1502)
    Not sure.
    In your (correct) answer to part (a), you say that (2 - 9x)^4 = 16 - 288x + 1944x^2 + ...

    So, in part (b), you should have that f(x) = (1 + kx)(2 - 9x)^4 = (1 + kx)(16 - 288x + 1944x^2 + ...)
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    (Original post by Pangol)
    In your (correct) answer to part (a), you say that (2 - 9x)^4 = 16 - 288x + 1944x^2 + ...

    So, in part (b), you should have that f(x) = (1 + kx)(2 - 9x)^4 = (1 + kx)(16 - 288x + 1944x^2 + ...)
    So should I find the 4th term?
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    (Original post by Anonymous1502)
    So should I find the 4th term?
    You only need to compare this with A - 232x + Bx^2, so you only need the first three terms.
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    (Original post by Pangol)
    You only need to compare this with A - 232x + Bx^2, so you only need the first three terms.
    How do I compare?
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    (Original post by Anonymous1502)
    How do I compare?
    If they are the same, then then have the same constant term, the same coefficient of x, the same coefficient of x^2, etc.
 
 
 
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