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.

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Basically, you have: $A = x^0+x^1+...+x^{n}$, and $xA = x^1+x^2+...x^{n+1}$

Now, $$A-xA = (x^0+x^1+x^2+...+x^{n})-(x^1+x^2+...x^{n+1}) = 1 - x^{n+1}$$ Also, $$A - xA = A(1-x)$$

So: $$A = \frac{1-x^{n+1}}{1-x}$$

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i'm even more lost now. How did you get that? – Jack F Oct 16 '12 at 23:16
can you show me step by step how to proof that problem – Jack F Oct 16 '12 at 23:18
@JackF clearer now? – Alex Oct 16 '12 at 23:19
yes thank you very much – Jack F Oct 16 '12 at 23:25

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 $$A=\frac{1-x^{n+1}}{1-x}\;.$$

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