So basically it's been over a decade since i used this stuff and now i could not remember even if my life depended on it...
The challenge is how to calculate $n$ when given the following formula $1.01^n + 1.01^{n-1} + 1.01^{n-2} + \cdots + 1.01 = 300$ ($p$ is a percentage $> 1$ for growth - so in this example it would be a growth of $1\%$ over a period of $n$ until $300$ is reached).
It's enough to point me in the right direction with a keyword I can search for, but if you have time on your hands and feel friendly today pls don't hold back with a full explanation :D
@Mods: Since I have no Idea where to place this Tag-wise feel free to move it where appropriate
If you prove that $\sum\limits_{k=0}^{n} p^k = \frac{1-p^{n+1}}{1-p}$ I think you can easily solve in $n$ the following : $\frac{1-p^{n+1}}{1-p} = 300$.
In order to prove the equality $\sum\limits_{k=0}^{n} p^k = \frac{1-p^{n+1}}{1-p}$ a simple way is to multiply both sides by $(1-p)$ and overlap to sum.
$$(1-p)\sum\limits_{k=0}^{n} p^k = \sum\limits_{k=0}^{n} p^k -p \sum\limits_{k=0}^{n} p^k$$
Overlap as follows
$$1 + p + p^2 + p^3 + \cdots + p^n$$
$$\hspace{1.7cm}-p-p^2-p^3-\cdots-p^{n}-p^{n+1}$$
From here you can see that if you sum all the terms cancel out except for the first and the last one, i-e $(1-p)\sum\limits_{k=0}^{n} p^k = 1-p^{n+1}$, which leads to what we wanted to prove if you divide for $(1-p)$
Note that I didn't consider the case where $p = 1$ since is a trivial case to consider.