This question is going to need some context. The TL;DR is:
How do I get a function equal to $\sum_{i=1}^n1.15^i$ without using sums?
I'm using Google sheets to keep track of something and as far as I know it doesn't support sequences or sums in one line (I don't want to have every terms of sequence in one cell and then sum the cells). Though this isn't a question about how to do things in Google sheets. Last time I had a sequence I managed to find a function using simple factors with Wolfram Alpha, equal to the sequence.
The first sequence is :
$A_n = A_{n-1}\times1.15$
$A_0 = a $
$A_0$ is variable and goes in a cell. For this sequence, I use $A_n = A_0\times1.15^n$, or as a function $f_a(n)=a\times1.15^n$.
The second sequence $B$ is the sum of all the terms of $A$.
$B_n = \sum_{i=1}^n A_i $
$ B_n = \sum_{i=1}^n A_0\times1.15^i$
$ B_n = A_0 \sum_{i=1}^n 1.15^i$
And that is problematic to calculate with a "google sheets friendly" function. For example, $B_3$ goes like this:
$B_3 = A_3 + A_2 + A_1 + A_0$
$B_3 = a\times1.15^3 + a\times1.15^2 + a\times1.15 + a$
$B_3 = a (1.15^3 + 1.15^2 + 1.15 + 1)$
So I have get this series of polynonial-looking terms and I have no idea how I can write a function that calculates $B_3$ and $B_n$ easily, without having to use a cell for each term. I've reduced my problem to calculating the sum I gave in the TL;DR but I'm stuck there.
If it's unclear what I'm asking, I just want to know if there's an easy way to calculate $\sum_{i=1}^n1.15^i$ for a given $n$ without using sums (i.e. without having to calculate each term of the sum). "No" is an acceptable answer, and any other way to express the same thing is welcome.
Also, is there a name for the act of finding an expression of $U_n$ that uses only $U_0$ instead of $U_{n-1}$?
The formula you want is $$\sum_{i=0}^n x^i = \frac{1-x^{n+1}}{1-x}.$$
This can be seen as follows:
\begin{align*} (1-x)(1 + x + \cdots + x^n) &= (1 + x + \dots + x^n) - x(1 + x + \dots + x^n)\\ &=(1 + x + \dots + x^n) - (x + x^2 + \dots + x^{n+1})\\ &= 1 - x^{n+1}. \end{align*}
Then plug in your $x = 1.15$ (and subtract $1$ since you begin with the index $i=1$ instead of $i=0$).