Calculating sum of consecutive powers of a number
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.
$\endgroup$ 14 Answers
$\begingroup$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.
$\endgroup$ 3 $\begingroup$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 $S_n$ $$ 1-P=\dfrac{1-P^{n+1}}{S_n}\\ S_n=\dfrac{1-P^{n+1}}{1-P} $$
$\endgroup$ 1 $\begingroup$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 $P-1$.
$\endgroup$ 0 $\begingroup$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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