How do you power represent $\frac{(1+x)}{(1-x)^2}$?

Question

$f(x)=\frac{(1+x)}{(1-x)^2}$ how do you power represent this formula?

I tried it by dividing it into $\frac{1}{(1-x)^2} + \frac{x}{(1-x)^2}$ but could not get a right answer.

Answer 1

Hint

$$\frac{1+x}{(x-1)^2}=\frac{x-1+2}{(x-1)^2}$$

Answer 2

The trick is to obtain $(1-x)^{-2}=\sum_{n\ge 0}(n+1)x^n$, either by the binomial formula or by differentiating $(1-x)^{-1}$. The final result is $\sum_{n\ge 0}(2n+1)x^n$.

Answer 3

By using $\frac{1}{1-x}= \sum_{n=0}^{\infty}x^n$ and the Cauchy product of Power Series you can get

$$\frac{1}{(1-x)^2} = \left( \frac{1}{1-x} \right)^2 =\left( \sum_{n=0}^{\infty}x^n \right)^2 = \sum_{n=0}^{\infty}(n+1)x^n $$ so $$ \frac{1}{(1-x)^2} + \frac{x}{(1-x)^2}=\sum_{n=0}^{\infty}(n+1)x^n+x \sum_{n=0}^{\infty}(n+1)x^{n+1} = \sum_{n=0}^{\infty}(2n+1)x^n$$