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Here is my problem, I want to compute the $$\sum_{i=0}^n P^i : P\in ℤ_{>1}$$ I know I can implement it using an easy recursive function, but since I want to use the formula in a spreadsheet, is there a better approach to this?

Thanks.

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  • $\begingroup$ Note to the casual reader: the way P to the power of i is rendered with smaller font sizes, makes it look like this is a lowercase p. But no, there is only one, capital P and this can be verified by zooming in with the browser. (I almost asked, what's the capital P vs. the small p?) $\endgroup$
    – Irfy
    Feb 5, 2019 at 22:35

4 Answers 4

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If we call the sum $S_n$, then $$P \cdot S_n = P + P^2 + P^3 + \cdots + P^{n+1} = S_{n} + (P^{n+1} - 1).$$

Solving for $S_n$ we find: $$(P - 1) S_n = P^{n+1} - 1$$ and $$S_n = \frac{P^{n+1}-1}{P-1}$$

This is a partial sum of a geometric series.

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    $\begingroup$ Wikipedia has the operands of the subtractions in the numerator and denominator reversed from this, $\frac{1-r^n}{1-r}$. Are they wrong or is this wrong or am I missing a difference that reconciles them? Link: en.m.wikipedia.org/wiki/Geometric_series $\endgroup$ Feb 7, 2019 at 3:00
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    $\begingroup$ Answering my own question: the numerator and the denominator always end up having the same sign. $\endgroup$ Feb 7, 2019 at 3:12
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    $\begingroup$ @JosephGarvin These two quantities are the same as $$\frac{1-r^n}{1-r} = \frac{(-1)(r^n-1)}{(-1)(r-1)} = \frac{r^n-1}{r-1}$$ $\endgroup$
    – Joel
    Mar 14, 2019 at 16:58
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We have $$\begin{array}{l} S_n&=1+P+P^2+P^3+\cdots+P^n\\ P\cdot S_n&=0+P+P^2+P^3+\cdots+P^n+P^{n+1} \end{array}$$ Subtracting two above equations gives $$ S_n-P\cdot S_n=1-P^{n+1} $$ divide by $S_n$ $$ 1-P=\dfrac{1-P^{n+1}}{S_n}\\ S_n=\dfrac{1-P^{n+1}}{1-P} $$

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  • $\begingroup$ Although other answers exist, I find this to be the most intuitive and easy to understand. Thanks! $\endgroup$
    – Gautam J
    Jul 10, 2022 at 9:13
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The elements of your sum follow a geometric rule. It happens that the sum of a geometric series has a simple formula (if $P$ is not $1$) :

$$\sum_{i=0}^n P^i = \dfrac{P^{n+1} -1}{P-1}$$

EDIT : Let's prove this !

$(P-1)(P^n +P^{n-1}+...+1)= (P^{n+1} -P^n) +(P^n -P^{n-1})+(P^{n-1}-P^{n-2}) +...+(P-1) = P^{n+1} -1$

You have the result by dividing both sides by $P-1$.

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That's a geometric series. There are $n+1$ terms starting from the first term of $P^0 = 1$, and the sum is given by $\displaystyle \frac{P^{n+1}-1}{P-1}$

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