Explicit formula for recursive geometric/arithmatic series In my Algebra 2 class, we have come upon a question that the class could not solve, and that the teacher has neglected to remove from the given packet for several years because of this. The problem is fairly easy to derive a tentative recursive formula from, being $a_1= 2000$, $a_n = (a_{n-1} + 2000) * 1.08$ (which is possibly incorrect based on the scenario).
The question is as follows (parenthesis indicate my notes):
Mr. and Mrs. Sanchez want to invest money for their child's college education. They have decided to invest $2000 at the beginning of the year, for the next 10 years. If the investment is in an account that earns 8% annual interest, compounded once per year (assumed at the end of the year, before the next deposit), how much will their investment be worth at the end of the 10 years (assumed to be after interest the 10th year)?
What possible explicit formula can be derived from this question? Is the recursive formula that we have derived correct?
Edit: as per the comments, I have realized that our interpretation of the recursive formula is incorrect. Rather, $a_1$ should be 2160 and $a_0$ should be 2000.

Let $u_n$ be the value of the investment after $n$ years have elapsed including the interest added at the end of that period, $n \geq 0$.
$$u_0 = 2000 \\
u_{n} = 1.08u_{n-1} + 2000, n > 0
$$
The solution is $$u_n = 1.08^n \times 27000 - 25000$$
$$u_{10} - 2000 = 1.08^{10} \times 27000 - 25000 - 2000 = 33291 - 2000 = 31291$$

How to get the formula from the recurrence:
We can find a $v_n$ that satisfies $v_n = 1.08v_{n-1}+2000$. Such a $v_n = -25000$.
Now what's $u_n - v_n$ where $u_n$ is the recurrence at the start. It's $u_n - v_n = 1.08(u_{n-1} - v_{n-1}) = 1.08^nC$ for some $C$.
So $u_n = 1.08^nC - 2000$. Since $u_0 = 2000$, we get $C = 27000$.
