<s>[INST] proof by by directly manipulating the sum: Prove that 
$$\sum_{i=0}^{n} x ^{i} = \frac{ 1-x ^{n+1} }{ 1-x }$$
to be used by  directly manipulating the sum: let  A be the sum, and show that xA = A + x^(n+1) -1
I don't get how its going to equal $\frac{ 1-x ^{n+1} }{ 1-x }$
$$xA=x\sum_{i=0}^n x^i=x(x^0+x^1+x^2+x^3+...)=x^1+x^2+x^3+x^4+...$$
so then i have 
$$\sum_{i=0}^{n} x ^{n+1}-1$$
I'm stock on how its going to equal one to each other.
 [/INST] Write it out like this:
$$\begin{align*}
A&=x^0+\color{red}{x^1+x^2+\ldots+x^n}\\
xA&=\quad\quad\,\color{red}{x^1+x^2+\ldots+x^n}+x^{n+1}\\
A-xA&=x^0+\qquad\qquad\color{red}{0}\qquad\quad\,-x^{n+1}
\end{align*}$$
Then $(1-x)A=A-xA=x^1-x^{n+1}=1-x^{n+1}$, so $$A=\frac{1-x^{n+1}}{1-x}\;.$$
</s>