Answer
$\frac{1+x}{1-x}=1+2x+2x^2+2x^3+...+x^n(x+1)+...$
Work Step by Step
The Taylor series for $\frac{1}{1-x}$ is
$1+x+x^2+x^3+...+x^n+...$
So, the Taylor series for $\frac{x}{1-x}$ is
$x+x.x+x.x^2+x.x^3+...+x.x^n...$
$x+x^2+x^3+x^4+...+x^{n+1}$
Hence,
$\frac{1}{1-x}+\frac{x}{1-x}$
$=1+x+x^2+x^3+...+x^n+...+(x+x^2+x^3+x^4+...+x^{n+1})$
$=1+2x+2x^2+2x^3+...+x^n(x+1)+...$
We are given $\frac{1+x}{1-x}=\frac{1}{1-x}+\frac{x}{1-x}$
so $\frac{1+x}{1-x}=1+2x+2x^2+2x^3+...+x^n(x+1)+...$
