# How do you go from the first step where the summation of (1-p)^{k-1} transform to the fraction

How do you go from the first step where the summation of $\left(1-p{\right)}^{k-1}$ transform to the fraction
$\sum _{k=1}^{n}p\left(1-p{\right)}^{k-1}=p\frac{1-\left(1-p{\right)}^{n}}{1-\left(1-p\right)}$
How to simplify the summation?
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Explanation:
For all $a\ne 1$, one can prove that $\sum _{k=0}^{n}{a}^{k}=\frac{1-{a}^{n+1}}{1-a},$ for all $n\in \mathbb{N}$
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Step 1
Firstly, lets get rid of p,which appears everywhere:
$\sum _{k=1}^{n}p\left(1-p{\right)}^{k-1}=p\sum _{k=1}^{n}\left(1-p{\right)}^{k-1}$
Now,lets denote $a=1-p$,where a is an integer.
Now denote with $S=\sum _{k=1}^{n}{a}^{k-1}$
Lets calculate the product between a and S:
$aS=\sum _{k=1}^{n}{a}^{k}$
$aS-S=\sum _{k=1}^{n}{a}^{k}-S$
Now lets subtract S from the sum:
$\left(a-1\right)S=\sum _{k=1}^{n}{a}^{k}-\sum _{k=1}^{n}{a}^{k-1}$
$\left(a-1\right)S={a}^{n}-1$
$S=\frac{{a}^{n}-1}{a-1}$
$S=\frac{1-{a}^{n}}{1-a}$
Step 2
Now all that we have to do is to replace a with $p-1$
$S=\frac{1-\left(1-p{\right)}^{n}}{1-\left(1-p\right)}$
$pS=p\frac{1-\left(1-p{\right)}^{n}}{1-\left(1-p\right)}$
$p\frac{1-\left(1-p{\right)}^{n}}{1-\left(1-p\right)}=pS=p\sum _{k=1}^{n}\left(1-p{\right)}^{k-1}=\sum _{k=1}^{n}p\left(1-p{\right)}^{k-1}$