Write the taylor expansion of $\displaystyle x/(1-x)^2$
I'm assuming you mean the Taylor series about x=0.
We know from the sum of a geometric series that $\displaystyle 1+x+x^{2}+x^{3}+ ... = \frac {1}{1-x} $
then $\displaystyle \frac{d}{dx} (1+x+x^{2}+x^{3}+ ...) = 1+2x+3x^{2}+4x^{3} +... = \frac{d}{dx} \frac{1}{1-x} = \frac{1}{(1-x)^{2}}$
so $\displaystyle \frac{x}{(1-x)^{2}} = x (1+2x+3x^{2}+4x^{3}+...) = (x+2x^{2}+3x^{3}+4x^{4} + ... ) = \sum_{n=1}^{\infty}nx^{n} $
The binomial series...
$\displaystyle (a+x)^{n} = a^{n} + \binom{n}{1} a^{n-1}x + \binom{n}{2} a^{n-2}x^{2} + \binom{n}{3} a^{n-3} x^{3} + \dots$ (1)
... in the case $\displaystyle a=1$ , $\displaystyle n=-2$ and changing $\displaystyle x$ into $\displaystyle -x$ becomes...
$\displaystyle (1-x)^{-2} = 1 + 2x + 3 x^{2} + 4 x^{3} + \dots$ (2)
If you multiply each term of (2) by $\displaystyle x$ you obtain what you have requested. That isn't the only way to arrive to the result...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$