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 \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&#039;m stuck after that.. Can I have a hint? I&#039;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?

 

[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?