What's wrong with my binomial expansion of $\frac{1}{(1-x)}$

Question

I am trying to use the binomial formula to expand the function $\dfrac{1}{1-x}$

$\dfrac{1}{1-x}=(1-x)^{-1}=1+(-1)x+(-1)(-2)\dfrac{x^2}{2!}+(-1)(-2)(-3)\dfrac{x^3}{3!}...$

The sign are all wrong they get much more similar to the series of $\dfrac{1}{1+x}$, which is an alternating series.

So what did I do wrong here?

Answer 1

Try replacing $x$ with $-x$ in your expansion since $1 - x = 1 + (-x)$.

Answer 2

Simply by long division you get $$\frac {1}{1-x} = 1+x+x^2+x^3+....$$