If x is some fixed real number, for each $ n \in N $ find the sum of $1-x-x^2-x^3-...-x^n$. If x is some fixed real number, for each $ n \in N $ find the sum of $1-x-x^2-x^3-...-x^n$.
The problem comes from "A friendly introduction to analysis" by Kosmala 2nd ed.
My approach to solving this problem was that I recognized that this looks similar to a geometric progression, so using the help of the internet, we have $1+r+r^2+r^3+...+r^n = \frac{1-r^{n+1}}{1-r}$. By rearranging the original expression, I can deduce that the summation of the expression is equal to $\frac{x^{n+1}-2x+1}{1-x}$ when $x \neq 1$ and $1-n$ when $x=1$. My question is, how would I go about attaining this answer had I not used the help/prior knowledge of the formula for the summation of the geometric progression?

You could have done it the way we derive the formula for geometric progressions:
$$S:= 1 - x - x^2 - \cdots - x^n$$
Then
$$Sx = x-x^2-x^3 - \cdots - x^{n +1}$$
Thus $$Sx - S = 2x - 1 - x^{n + 1} \implies S = \left\{\begin{array}{cc}\frac{x^{n + 1} - 2x + 1}{1 - x},&  \mbox{ if } x \neq 1\\
1- n,&  \mbox{ if } x = 1\end{array}\right.$$
