I am trying to make a simple model for calculating the time for a buss to finish a whole lap. Assumptions of my model: all stops are evenly spaced and it takes the time $t_0$ for the buss to travel between each stop, the velcoity is consant. Before the first stop the buss will also travel a time $t_0$. At each stop the amount of travelers is increased by $x$ persons per second and lets say a person takes the time $k$ seconds to enter the bus and take a seat.
To reach the first stop the buss will travel the time $t_0$, and the time that will have been spent on the first stop would be $t_0kx$. The buss will then travel to the second stop and that would take the time $t_0$ but now the time that will be spent on the second stop will be $(t_0+t_0kx +t_0)kx$ because $t_0+t_0kx +t_0$ is the time that the bus has been travelling yet, and during all that time the amount of customers have incresed at the second bus stop. The sum to find the total time:
\begin{align} T=&t_0 + t_0kx\\ +&t_0 +(t_0 + t_0kx + t_0)kx\\ +&t_0 +(t_0 + t_0kx + t_0 +(t_0 + t_0kx + t_0)kx + t_0)kx\\ &.\\ &.\\ &. \end{align} I found this recursive formula: \begin{align} f_n=t_0+(t_0+\sum_{i=1}^{n-1}f_i)kx, \;\;\;\;\;\; f_1=t_0+t_0kx \end{align}
My main question is if there exists a general formula for the sum and what it would be. Also if there are any suggestions for a better model I would also be thankful!
By expansion,
$$f_1 = t_0 + t_0 kx = t_0(1 + kx)$$ $$\begin{align}f_2 &= t_0 + (t_0 + t_0 + t_0 kx)(kx) \\ &= t_0 + 2 t_0 kx + t_0 (kx)^2 \\ &= t_0(1 + 2 kx + (kx)^2) \\ &= t_0(1 + kx)^2 \end{align}$$ $$\begin{align} f_3 &= t_0 + (t_0 + t_0 (1 + kx) + t_0(1 + 2 kx + (kx)^2))(kx) \\ &= t_0 + (3 t_0 + 3 t_0 kx + t_0 (kx)^2)(kx) \\ &= t_0 + 3 t_0 kx + 3 t_0 (kx)^2 + t_0 (kx)^3 \\ &= t_0 (1 + 3 kx + 3 (kx)^2 + (kx)^3) \\ &= t_0 (1 + kx)^3 \end{align}$$ $$\begin{align} f_4 &= t_0 + (t_0 + t_0 (1 + kx) + t_0(1 + 2 kx + (kx)^2) + t_0 (1 + 3 kx + 3 (kx)^2 + (kx)^3))(kx) \\ &= t_0 + (4 t_0 + 6 t_0 kx + 4 t_0 (kx)^2 + t_0 (kx)^3)(kx) \\ &= t_0 + 4 t_0 kx + 6 t_0 (kx)^2 + 4 t_0 (kx)^3 + t_0 (kx)^4 \\ &= t_0 (1 + 4 kx + 6 (kx)^2 + 4 (kx)^3 + (kx)^4) \\ &= t_0 (1 + kx)^4 \end{align}$$ Therefore, the general formula is $$f_n = t_0 (1 + kx)^n$$ which can be proven by induction.
To find $\sum_{n = 1}^m f_n$, we can use the geometric sum formula. $$\begin{align} \sum_{n = 1}^m f_n &= \sum_{n = 1}^m t_0 (1 + kx)^n \\ &= t_0 (1 + kx) \sum_{n = 1}^m (1 + kx)^{n - 1} \\ &= t_0 (1 + kx) \frac{1 - (1+kx)^m}{1 - (1+kx)} \\ &= t_0 \frac{1 + kx}{kx}((1+kx)^m-1) \end{align}$$