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Can someone explain to me how the following summation goes from the left to the right:

$\sum_{i=1}^{r}n(1-p)^{i-1} = \frac{n(1-(1-p)^r)}{p}$.

I have used the formula for a Geometric series. My common ratio is: $(1-p)$. I get the following:

$\sum_{i=1}^{r}n(1-p)^{i-1} = n\cdot\frac{1-(1-p^r)}{1-(1-p)}$.

From here, I simply don't know how to get to $\frac{n(1-(1-p)^r)}{p}$

GoldenRetriever
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  • For geometric sequence, all you need to know is the first term $b$ and the common ratio $q$. –  Jan 14 '21 at 16:00
  • i don't think the infinite series should be used here. @mrsamy, how would you get to the final answer i have posted by knowing these terms? – GoldenRetriever Jan 14 '21 at 16:11
  • In your question, $n$ is a constant, which has nothing to do with $i$. The first implication in what you have done is incorrect. –  Jan 14 '21 at 16:16
  • I don't understand what you mean? could you elaborate a bit more? I have just used a simple formula for geometric series. – GoldenRetriever Jan 14 '21 at 16:18
  • @GoldenRetriever I wrote an answer. Let me know if you have more questions. –  Jan 14 '21 at 16:21

2 Answers2

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In your question, $n$ is a constant, which has nothing to do with $i$. The first implication of what you have done is incorrect.

Note that $n(1-p)^{i-1}$ means $n\cdot ((1-p)^{i-1})$, and

$$ \sum_{i=1}^{r}n(1-p)^{i-1}=n\sum_{i=1}^{r}(1-p)^{i-1} $$

all you need to work out is to find a formula for $$ \sum_{i=1}^{r}x^{i-1} $$ and then set $x=1-p$.

  • you are right. I did a mistake with n. But when I use the formula as posted, I don't see how I get to the final result I am looking for. – GoldenRetriever Jan 14 '21 at 16:26
  • @GoldenRetriever: that is because you make another mistake; you applied the formula incorrectly. Do you know how to find a formula for $\sum_{i=1}^rx^{i-1}$? –  Jan 14 '21 at 16:37
  • @GoldenRetriever you are almost there: $$\sum_{i=1}^{r}n(1-p)^{i-1} = n\cdot\frac{1-(1-\color{red}{p}^r)}{1-(1-p)}$.$$ $p$ should be $(1-p)$. You just applied the formula incorrectly. –  Jan 14 '21 at 16:41
  • It's still not the result? see link: https://www.wolframalpha.com/input/?i=n*%28%281-1-%281-p%29%5Er%29%2F%281-1-%281-p%29%29%29 – GoldenRetriever Jan 14 '21 at 16:45
  • Where did I applied the formula incorrectly? – GoldenRetriever Jan 14 '21 at 16:54
  • @GoldenRetriever: As I said you should first find a formula for $\sum_{i=1}^r x^{i-1}$. Do you know how to find that? –  Jan 14 '21 at 16:59
  • I would need a series for that? wouldn't I? Btw. I just found out that $n\cdot\frac{1-(1-p^r)}{1-(1-p)}$ is in fact $ \frac{n(1-(1-p)^r)}{p}$. Now I just need to figure out how. Simple elimination i guess. – GoldenRetriever Jan 14 '21 at 17:02
  • @GoldenRetriever: you are thinking too hard. it is just $1-(1-p)=p$ in the denominator. –  Jan 14 '21 at 17:03
  • haha I guess that's what happens when you spend too much time on a thing. Thanks man! – GoldenRetriever Jan 14 '21 at 17:06
  • @GoldenRetriever: no. $1-(1-p)=1-1+\color{red}{p}=p$. –  Jan 14 '21 at 17:07
  • thanks ! Speaking about the finding of $x^{i-1}$. Why do you have to do that when I just used the geometric formula? – GoldenRetriever Jan 14 '21 at 17:08
  • @GoldenRetriever: you are welccome. That is just saving your writing when you write $x$ instead of $1-p$ everywhere. –  Jan 14 '21 at 17:09
  • but why did you move out the n? Are you supposed to do that when it's a Geometric sequence? – GoldenRetriever Jan 14 '21 at 17:11
  • @GoldenRetriever: you don't have to. That simplifies things. Very much like when you calculate $35\cdot 3+35\cdot 3+35\cdot 4$, you would like to factor out $35$ first. –  Jan 14 '21 at 17:12
  • ah okay. I hope to see you answer my upcoming questions. thanks again. have a good day. – GoldenRetriever Jan 14 '21 at 17:14
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\begin{align*} \sum_{i=1}^{r}n(1-p)^{i-1} & =n\sum_{i=0}^{r-1}(1-p)^i =[\text{\(r\) terms}] =n\cdot\frac{1-(1-p)^r}{1-(1-p)} =\frac{n\bigl(1-(1-p)^r\bigr)}{p} \end{align*}

mf67
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