Binomial Expansion: Coeff. of x^n in (1+x)^n/(1+2x)^2

[whkoh]

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

 

[HallsofIvy]

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