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    I probably could do this...the really long way...but when this question says 'as a product of primes' - I have no idea!!


    a: By considering the general term, find the coefficient of the term x^14 in the expansion of:

    [x^3 + (3/x^2)]

    expressing your answer as a product of primes.

    b: Write down the expansion of (1 + x)^6. Hence by letting x= z + z^2 expand (1 + z + z^2) in ascending powers of z as far as the term in z^4

    or just look at the attachement.

    Please help!
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    Work out what term in the expansion will give you x^14. Start from (x^3)^9 and (3/x^2)^9 as halfway, and work from there.

    Then use nCr where n is 18 and r is the power you found above

    Break your co-efficient down into its prime factors

    b is pretty straightforward. Expand, substitute and stop when the powers hit z^4. Simplify
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    (Original post by Trangulor)
    Work out what term in the expansion will give you x^14. Start from (x^3)^9 and (3/x^2)^9 as halfway, and work from there.

    Then use nCr where n is 18 and r is the power you found above

    Break your co-efficient down into its prime factors

    b is pretty straightforward. Expand, substitute and stop when the powers hit z^4. Simplify
    I really, really appreciate your help...but now I'm even more confused :confused:
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    Could you/anyone clarify
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    Ok. The standard binomial is:

    (a+b)^n=a^n + na^(n-1) b + 0.5n(n-1)a^(n-2) b^2 + ... + b^n

    Firstly, ignore the co-efficients and just concentrate on the powers. How can you get x^14 from (x^3)^p times (3/x^2)^q, where p+q=18?

    Spoiler:
    Show
    By laws of indices, x^3p times x^-2q = x^14

    3p-2q=14

    2p + 2q = 36 (double p+q=18)

    Add: 5p=50

    p=10, q=8

    with these numbers, work out the x^14 term and its co-efficient, which you need to write as its prime factors
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    (Original post by Trangulor)
    Ok. The standard binomial is:

    (a+b)^n=a^n + na^(n-1) b + 0.5n(n-1)a^(n-2) b^2 + ... + b^n

    Firstly, ignore the co-efficients and just concentrate on the powers. How can you get x^14 from (x^3)^p times (3/x^2)^q, where p+q=18?

    Spoiler:
    Show
    By laws of indices, x^3p times x^-2q = x^14

    3p-2q=14

    2p + 2q = 36 (double p+q=18)

    Add: 5p=50

    p=10, q=8

    with these numbers, work out the x^14 term and its co-efficient, which you need to write as its prime factors
    ok now your gunna think that im a complete idiot! ive never seen that formula - could you write it down in an attachment becuase i dont understand whats going on.

    also i dont seem to understand what the answer is:confused:
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    Bah, edexcel website is down

    The second one may be the more familiar, requires the nCr button

    Edit: The formula is in the Edexcel formula booklet, so when the website is back up, google for it. Might want to give it a go now, as it may just be my internet playing up
    Attached Images
     
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    (Original post by Trangulor)
    Bah, edexcel website is down
    Here are the formulae
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    Thanks Steve

    NM, the formula is on Page 7, so you won't need to remember it
 
 
 
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