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Maths binomials question quick help?? watch

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    This is the question:


    In the expansion of (1+3x)^n the coefficient of x^3 is double the coefficients of x^2.work out the value of n.
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    (Original post by georgee913)
    This is the question:


    In the expansion of (1+3x)^n the coefficient of x^3 is double the coefficients of x^2.work out the value of n.
    and your working?
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    Try writing out the binomial expansion up to x^3, use the n!/(r!(n-r)!) equation for nCr. Then equate coefficients.

    Alternatively, use this formula if you're familiar with it : http://www.physics.udel.edu/~watson/...binomial20.gif
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    I got up to this point but i just need help simplifying this
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    n! / (n - 2)! = n(n - 1). Similar reasoning for n! / (n - 3)!
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    (Original post by georgee913)
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    I got up to this point but i just need help simplifying this
    the answer is 8, but i will give you then next step of working and let you figure the rest out.
    phrase the question like this: Ax^3=2Bx^2, and then:
    2 times (n(n-1)(n-2)/2(n-2)! all times (3x)^2) is equal to n(n-1)(n-2)(n-3)/3!(n-3)! all times (3x)^3.
    then simplify and turn into a cubic
    do not forget that n!= n(n-1)(n-2).... write this on the numerator instead of n! to make life easier, however only do n up to the power of x eg x^3, then n!= n(n-1)(n-2)(n-3)
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    (Original post by Anonymouse 921)
    the answer is 8, but i will give you then next step of working and let you figure the rest out.
    phrase the question like this: Ax^3=2Bx^2, and then:
    2 times (n(n-1)(n-2)/2(n-2)! all times (3x)^2) is equal to n(n-1)(n-2)(n-3)/3!(n-3)! all times (3x)^3.
    then simplify and turn into a cubic

    I got 8 at first but the mark scheme said that n is 4 ::

    (1 + 3x) n = 1 + n C 1 (3x) + n C 2 (3x) 2 + n C 3 (3x) 3 + … n – 2 = 2 so n = 4
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    (Original post by georgee913)
    I got 8 at first but the mark scheme said that n is 4 ::

    (1 + 3x) n = 1 + n C 1 (3x) + n C 2 (3x) 2 + n C 3 (3x) 3 + … n – 2 = 2 so n = 4
    i realised my mistake, I did not include the coefficient of x^2 and X^3 at the beginning, give me 5 mins while I redo the question
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    to be picky you wrote (1)n in your working... it should be (1)n-2 or (1)n-3

    although the value is the same....
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    (Original post by georgee913)
    I got 8 at first but the mark scheme said that n is 4 ::

    (1 + 3x) n = 1 + n C 1 (3x) + n C 2 (3x) 2 + n C 3 (3x) 3 + … n – 2 = 2 so n = 4
    new working:
    ( a full stop is equal to times, I do it to avoid confusion with x)
    2.3^2.(n(n-1)/2)=3^3.(n(n-1)(n-2)/3!)

    this simplifies to 2n^2-2n=n^3-3n^2+2n, or 0=n^3-5n^2+4n, where this factorises equals n(n-4)(n-1), and n cannot be 1 or zero, then it must equal 4
 
 
 
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