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    Hi TSR, needed some help with this question:

    (3-2x)^-1

    I have to expand to get the coefficients of the x term and x^2 term, and to also state the range of x for which it is valid

    Am I right to begin by taking (3-2x)^1 as 3(1-(2/3)x)^-1 so I can apply the binomial negative exponent formula?

    edit: using this method I got 3+2x+(4/3)x^2, is this correct?
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    (Original post by eternaforest)
    Hi TSR, needed some help with this question:

    (3-2x)^-1

    I have to expand to get the coefficients of the x term and x^2 term, and to also state the range of x for which it is valid

    Am I right to begin by taking (3-2x)^1 as 3(1-(2/3)x)^-1 so I can apply the binomial negative exponent formula?

    edit: using this method I got 3+2x+(4/3)x^2, is this correct?
    The 3 at the front should also be to the power of -1, i.e. it should be 1/3.
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    (Original post by tiny hobbit)
    The 3 at the front should also be to the power of -1, i.e. it should be 1/3.
    Ahh thank you, silly me

    I will try it again
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    (Original post by eternaforest)
    Hi TSR, needed some help with this question:

    (3-2x)^-1

    I have to expand to get the coefficients of the x term and x^2 term, and to also state the range of x for which it is valid

    Am I right to begin by taking (3-2x)^1 as 3(1-(2/3)x)^-1 so I can apply the binomial negative exponent formula?

    edit: using this method I got 3+2x+(4/3)x^2, is this correct?
    If I'm reading you correctly, it looks like you've taken out the factor of 3 incorrectly.

    \displaystyle \frac{1}{3-2x} = \frac{1}{3 (1 - (2/3)x)} = \frac{1}{3} \left(\frac{1}{1 - (2/3)x}\right)

    Can you take it from here?
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    (Original post by Gregorius)
    If I'm reading you correctly, it looks like you've taken out the factor of 3 incorrectly.

    \displaystyle \frac{1}{3-2x} = \frac{1}{3 (1 - (2/3)x)} = \frac{1}{3} \left(\frac{1}{1 - (2/3)x}\right)

    Can you take it from here?
    yeah someone pointed out before, also I have to do the binomial expansion like (1+x)^a = 1 + ax + a(a-1)x^2/2! + a(a-1)(a-2)x^3/3! ...

    I get 1/3 + 2/9x + 4/9x^2, is this right?
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    (Original post by eternaforest)
    I get 1/3 + 2/9x + 4/9x^2, is this right?
    Not quite - the term for x^2 isn't correct. Remember that

    \displaystyle \frac{1}{1-u} = 1 + u + u^2 + u^ 3 + \ldots

    then set  u = (2/3)x and remember to divide by that three!
 
 
 
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