Taylor Series expansion of $\frac{(1+x)}{(x-1)^2}$

Question

Show that the taylor Series expansion of $\frac{(1+x)}{(x-1)^2} = \sum_{n=0} (2n+1) x^n$ valid for $|x| < 1$

Since, $\frac{1}{(x-1)(x-1)} (1+x),$

According to what I learnt, I'm trying to prove this taylor expansion true by using the taylor series expansion of $\frac{1}{1-x} = \sum_{n=0} x^n $

I can only solve it by using this method but I have no clear direction and unsure how to prove it using $\frac{1}{1-x} = \sum_{n=0} x^n $

Answer 1

The advice is here $$ \begin{align*} \frac{1}{(1-x)^2} &= \left( \frac{1}{1-x} \right)^2 \\ &= \left( \sum_{n=0}^\infty x^n \right)^2 \\ &= \sum_{n=0}^\infty \sum_{k=0}^n x^n \\ &= \sum_{n=0}^\infty (n+1) x^n. \end{align*} $$

Answer 2

Another explanation could be $${1\over (1-x)^2}={d\over dx}{1\over 1-x}={d\over dx}\sum_{n=0}^\infty x^n=\sum_{n=1}^\infty nx^{n-1}=\sum_{n=0}^\infty (n+1)x^n$$