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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.

2 Answers 2

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