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I have a heard time seeing why is this true $\frac{1-y^n}{1-y}=(1+y+y^2+...+ y^{n-1})$

Could you show me some kind of proof, or an identity that would me to find this?

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    $\begingroup$ Have you tried multiplying through by the denominator? $\endgroup$
    – Mark S.
    Jul 22 '16 at 19:50
  • $\begingroup$ You can try polynomial division and notice a pattern. The most natural way to see it, IMO, is through induction because it's clearly true for $n=1,2,3,4$. $\endgroup$
    – user258700
    Jul 22 '16 at 19:52
  • $\begingroup$ A word you may want to learn is telescoping sum it means that in a sum each term and the next cancel. That is what happens for this sum if you multiply each side with $1-y$. $\endgroup$ Jul 22 '16 at 20:04
  • $\begingroup$ Covered by answers to this question, listed in Meta's List of generalizations of common questions. $\endgroup$
    – ccorn
    Jul 22 '16 at 20:21
  • $\begingroup$ Also answered by Proving the geometric sum formula by induction. $\endgroup$
    – ccorn
    Jul 22 '16 at 20:35
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This is a classic result for the sum $S_{n-1}$ of a geometric progression :

Let $$S_{n-1} = \sum_\limits{k=0}^{n-1}y^k=1+y+y^2+...+ y^{n-2}+ y^{n-1}$$

Thus, you have : \begin{align} y\times S_{n-1}&=y+y^2+y^3+...+y^{n-1}+ y^{n}\\ & \Rightarrow S_{n-1}-yS_{n-1} =1-y^n\\ & \Rightarrow S_{n-1}(1-y)=1-y^n\\ & \Rightarrow S_{n-1} = \frac{1-y^n}{1-y}\\ \end{align} $\square$

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$(1-y)(1+y+y^2+\ldots+y^{n-1})=1+ \not\!y+\not\!y^2+\ldots+\not\!y^{n-1}-\not\!y-\not\!y^2-\ldots-\not\!y^{n-1}-y^n=$ $$1-y^n$$

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You can check pretty easily that $1-y^n = (1-y)(1 + y + y^2 + \cdots y^{n-1})$. Notice that, in distributing, you get a positive copy of every monomial in $1 + y + \cdots + y^{n-1}$, and a negative copy of a ton of powers of $y$.

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We can proceed by induction on $n$. If $n=1$, we have $$\frac{1-y}{1-y}=1=\sum_{k=0}^0 y^k.$$ Now suppose that $$\frac{1-y^n}{1-y}=1+y+y^2+...+ y^{n-1}$$ For some $n\in\mathbb{N}$. Then we have $$ 1+y+y^2+...+ y^{n-1}+y^n= \frac{1-y^n}{1-y}+y^n=\frac{1-y^n}{1-y}+\frac{y^n(1-y)}{1-y}=\frac{1-y^{n+1}}{1-y}.$$ Therefore, the formula is valid for all $n\in\mathbb{N}$.

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This isn't a proof, more of a motivator to give you the intuition that the formula should work: calculate a few examples with $y=10$ and smallish $n$, after multiplying numerator and denominator by $-1$ to get

$$\frac{y^n-1}{y-1}$$

Example with $n=5$:

$$\frac{10^5-1}{10-1} = \frac{99999}{9} = 11111 = 10^4+10^3+10^2+10^1+10^0$$

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$$\begin{align} \frac{1-y^n}{1-y} &=\frac{1\color{red}{-y+y}\color{blue}{-y^2+y^2}-y^3+\cdots+y^{n-2}\color{magenta}{-y^{n-1}+y^{n-1}}-y^n}{1-y}\\ &=\frac{(1\color{red}{-y})+y(\color{red}{1}\color{blue}{-y})+y^2(\color{blue}{1}-y)+\cdots+y^{n-2}(1\color{magenta}{-y})+y^{n-1}(\color{magenta}{1}-y)}{1-y}\\ &=\frac{(1-y)(1+y+y^2+\cdots+y^{n-2}+y^{n-1})}{1-y}\\ &=1+y+y^2+\cdots+y^{n-2}+y^{n-1} \end{align}$$

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More generally, if

$$ s_n = \sum_{k=0}^{n-1} ar^k = as + ar + ar^2 + \dots + ar^{n-1}, $$

then

$$ s_n(1 - r) = s_n - rs_n = a - ar^n = a(1-r^n), $$

so

$$ s_n = \frac{a(1-r^n)}{1-r}. $$

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