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    • Thread Starter

    I've done the first part of the question but can't figure out the second part.

    Q - Hence, or otherwise, find the first 3 terms in the expansion of (x + 8) / sqroot(4 - 3x) as a series in ascending powers of x

    From part a I expanded sqroot(4 - 3x) to:

    1/2 + 3x/16 + 27x^2/256

    I'm unsure on what I do with the (x + 8)

    I tried expanding it and got:

    8^1(1 + x/8)^1 however on the 3rd term n-1 would equal 0 so I'm not sure how I expand (x + 8) if that's what I need to do.

    All help much appreciated
    • PS Helper

    PS Helper
    Urm... when you expand (x+8)^1 you get x+8 :p: But this isn't your problem.

    You can re-write your expression as (x+8)(4-3x)^{-\frac{1}{2}}, which in turn is equal to 3^{-\frac{1}{2}}(x+8)(\frac{4}{3}-x)^{-\frac{1}{2}}.

    What you need to do is expand (\frac{4}{3}-x)^{-\frac{1}{2}} (the first 3 terms will do), and then multiply the result by x+8. And then multiply that by 3^{-\frac{1}{2}}.
    • Thread Starter

    Aha I see, thanks!
    Yeah the first part was 1 / sqroot(4-3x). forgot to say that
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Updated: April 13, 2011

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