As $\frac1{(1-x)^2} = {1+2x+3x^2 +...nx^{n-1}} + \frac{x^n}{(1-x)^2} + \frac{nx^n}{(1-x)}$, is this is a function of only $x$ or both $x$ and $n$? $$\frac1{(1-x)^2} = {1+2x+3x^2 +...nx^{n-1}} + \frac{x^n}{(1-x)^2} + \frac{nx^n}{(1-x)}$$
Now, the LHS seems to be a function of only $x$, whereas the RHS seems to be a function of both $x$ as well as $n$. Please remove this ambiguity for me as I don't understand why is it that by simply choosing a value of $x$ on LHS is immaterial of what value $n$ holds.

This is equality ,not a function
$$\frac1{(1-x)^2} = {1+2x+3x^2 +...nx^{n-1}} + \frac{x^n}{(1-x)^2} + \frac{nx^n}{(1-x)^2}$$
$$\frac1{(1-x)^2} = \frac{d}{dx}\{x+x^2+x^3 +\cdots x^{n}\} + \frac{x^n}{(1-x)^2} + \frac{nx^n}{(1-x)^2}$$
$$x+x^2+x^3 +\cdots x^{n}=\frac{x(1-x^n)}{1-x}$$
$$\frac1{(1-x)^2} = \frac{d}{dx}\left\{\frac{x(1-x^n)}{1-x}\right\} + \frac{x^n}{(1-x)^2} + \frac{nx^n}{(1-x)^2}$$
$$\frac1{(1-x)^2}=\frac1{(1-x)^2}$$
This hold true for $$\forall \space  x \in \mathbb{R} -\{1\} $$
