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.
 A: 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 $P-1$.
A: 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.
A: 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 $S_n$
$$
1-P=\dfrac{1-P^{n+1}}{S_n}\\
S_n=\dfrac{1-P^{n+1}}{1-P}
$$
A: 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 $\displaystyle \frac{P^{n+1}-1}{P-1}$
