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Binomial expansion

  1. Jan 10, 2006 #1
    By writing [itex](1+x)[/itex] as [tex]\frac{1}{2}\left[1+\left( 1+2x\right) \right][/tex] or otherwise, show that the coefficient of [itex]x^n[/itex] in the expansion of [itex]\frac{\left(1+x\right)^n}{\left(1+2x\right)^2}[/itex] in ascending powers of x is [itex]\left(-1\right)^n\left(2n+1\right)[/itex].
    --
    I've tried expressing [itex](1+x)^n[/itex] as [tex]\left(\frac{1}{2}\right)^n \left[1+\left(1+2x\right)\right]^n[/itex], so as factor out (1+2x)
    [tex]\frac{\left(1+x\right)^n}{\left(1+2x\right)^2}[/tex]
    [tex]= \frac{\left({\frac{1}{2}}\right)^n \left[1+\left(1+2x\right)\right]^n}{\left( 1+2x \right)^2}[/tex]
    but I'm stuck after that.. Can I have a hint? I'm not sure how to write the general term of [itex]\left[1+\left(1+2x\right)\right]^n[/itex]: I get [itex]\frac{n!}{n!}\left(-1\right)^n x^n[/itex]
    Can someone point me in the right way?
     
    Last edited: Jan 10, 2006
  2. jcsd
  3. Jan 10, 2006 #2

    HallsofIvy

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    Have you considered letting u= 1+ 2x so that
    [tex]\left(\frac{1}{2}\right)^2\frac{\left(1+\left(1+2x\right)\right)^n}{\left(1+2x\right)^2}= \left(\frac{1}{2}\right)^2\frac{\left(1+u\right)^n}{u^2}[/tex]
    and then using the binomial theorem for the numerator?
     
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