# Sum of n with factoral exponent

I'm not sure what this is called, but the application is in counting the number of nodes in a consistently branching structure.

For example, $$5$$ nodes branch into $$5$$ nodes each, each branching again, etc, $$5$$ times overall.

I think the math would be:

$$5^1 + 5^2 + 5^3 + 5^4 + 5^5 = 3905$$

I'm just looking for a better way of calculating this. I looked up exponent rules, and they did not cover this scenario.

• Have you ever heard anything about geometric progressions? – AlessioDV Mar 13 at 15:03

This is known as a geometric series. Look up Geometric Series. Specifically if you want to find the sum of $$a+ar+ar^2+\cdots+ar^{n-1}$$, the formula is as follows: $$\sum_{k=0}^{n-1}ar^k=a\left(\dfrac{1-r^n}{1-r}\right)$$ where $$r$$ is the common ratio of the Geometric Progression.
For the specification you have mentioned, you just have to put in $$\begin{bmatrix}a \\ r\end{bmatrix}=\begin{bmatrix}5 \\ 5\end{bmatrix}$$.
Let $$q\in \mathbb{R}$$, $$q\neq 1$$, and $$n\in \mathbb{N}$$. It holds that $$q+q^2+\ldots + q^n = \frac{q}{1-q}(1-q^n).$$
Proof. Let $$S_n=q+q^2+\ldots + q^n$$. Then $$qS_n=q^2+\ldots+q^{n+1}$$ and $$S_n-qS_n=q-q^{n+1}=q(1-q^n),$$ from which it is easy to conclude. $$\Box$$