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    Hi guys,

    I am struggling with this question:

    (x^3-(1/2x))^12

    I need to find the first 4 terms...

    I have started by doing 12 (x^3)^12 + 12(x^3)^11(1/2x) etc. but I'm not sure if this is correct.
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    (Original post by Mazza2000)
    Hi guys,

    I am struggling with this question:

    (x^3-(1/2x))^12

    I need to find the first 4 terms...
     \displaystyle (a+b)^n = \binom{n}{0}a^nb^0 + \binom{n}{1}a^{n-1}b^1 + \binom{n}{2}a^{n-2}b^2 + ...

    so let n=12
    a = ?
    b = ?
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    (Original post by Mazza2000)
    Hi guys,

    I am struggling with this question:

    (x^3-(1/2x))^12

    I need to find the first 4 terms...
    The question is rather incomplete. It asks you to work out the first four terms - what does this mean? There is no general convention about what the first four terms are. It should really specify if it is the first four terms in ascending powers of  x that it wants or the first four descending powers of  x that it wants.

    Nonetheless, what have you tried so far?

    What is the binomial expansion of  \left(x+y\right)^{12} ? Work out the first four (ascending/descending) terms of that and substitute into your expansion  x = x^3, y = -\displaystyle \frac{1}{2} x .
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    i'm doing binomial expansion too and i think you must simply use the binomial theorem

    12C0 * (x^3*12) + 12C1 * (x^3*11)*(-1/2x) + 12C2 * (x^3*10)*(-1/2x)^2 + 12C3*(x^3*9)(-1/2x)^3....
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    (Original post by crashMATHS)
    The question is rather incomplete. It asks you to work out the first four terms - what does this mean? There is no general convention about what the first four terms are. It should really specify if it is the first four terms in ascending powers of  x that it wants or the first four descending powers of  x that it wants.

    Nonetheless, what have you tried so far?

    What is the binomial expansion of  \left(x+y\right)^{12} ? Work out the first four (ascending/descending) terms of that and substitute into your expansion  x = x^3, y = -\displaystyle \frac{1}{2} x .
    So I have started by --->This is meant to show 12 choose 0 (12/0)(x^3)^12 + (12/1)(x^3)&11(1/2x)+(12/2)(x^3)^10(1/2x)^2+(12/3)(x^3)^9(1/2x)^3
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    Ahh sorry I was just confused if what I was doing was correct or not but I've found the answer and it seems to be correct.
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    (Original post by Mazza2000)
    So I have started by --->This is meant to show 12 choose 0 (12/0)(x^3)^12 + (12/1)(x^3)&11(1/2x)+(12/2)(x^3)^10(1/2x)^2+(12/3)(x^3)^9(1/2x)^3
    You need to multiply by  \displaystyle - \frac{1}{2} x . Notice you have a negative in your expression. You are expanding  \displaystyle \left(x^3+\frac{1}{2} x \right)^{12} .

    Also, I get the impression from your edited OP you think  \displaystyle \binom{12}{0} = 12 . It doesn't - it equals to 1!
 
 
 
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