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I know that $\sum_{i=1}^n p^i = \frac{p-p^{n+1}}{1-p}$, but I am not sure how the i+1 factors into the closed form for $\sum_{i=1}^n p^{i+1}$, what is the closed form for the second sum?

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  • $\begingroup$ $\sum_{i=1}^n p^{i+1}=p \sum_{i=1}^n p^{i}$, But your formula is wrong. $\sum_{i=1}^n p^{i}=\frac{p-p^{n+1}}{1-p}$ $\endgroup$ Nov 3 '20 at 19:46
  • $\begingroup$ Thank you I will update it, so the +1 doesn't change the closed form? $\endgroup$
    – JGoss
    Nov 3 '20 at 19:49
  • $\begingroup$ It changes... by muliplying it by $p$ $\endgroup$ Nov 3 '20 at 19:53
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    $\begingroup$ so it is $\frac{p^2 - p^{n+2}}{1-p}$ ? $\endgroup$
    – JGoss
    Nov 3 '20 at 19:55
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    $\begingroup$ thank you I will create an answer for the question. $\endgroup$
    – JGoss
    Nov 3 '20 at 19:58
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You basically answered your question yourself.

You know, $\sum_{i=1}^n p^i = \frac{p \left(p^n-1\right)}{p-1}$

Now

$$\sum_{i=1}^n p^{i+1} = p \sum_{i=1}^n p^i$$

So you get $$\sum_{i=1}^n p^{i+1} = \frac{p^2 \left(p^n-1\right)}{p-1}$$

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    $\begingroup$ @hardmath fixed it $\endgroup$
    – user824530
    Nov 3 '20 at 20:10

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