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.

$\color{blue}{\text{Proof Using Discounting and Compounding}}$
Assume that we have a cash flow of $C$ at the beginning of each year for years $1$ to $n$, and that interest rate (or discount rate) is $r$. 
The present value PV (i.e at year $i=0$) of the cash flows is given by discounting each cash flow back to year $0$, i.e.
$$\text{PV}=\sum_{i=1}^n \frac C{(1+r)^i}=\frac Cr\left[1-\frac 1{(1+r)^n}\right]$$
(see below for proof)
The value required by the question is the future value FV at year $i=n+1$ which can easily be calculated by compounding PV by $n+1$ years, i.e. 
$$\text{FV}=\text{PV}(1+r)^{n+1}=\frac {C(1+r)}r\left[(1+r)^n-1\right]$$
Substituting $C=2000, r=0.08, n=10$ gives the value of $\color{red}{31,291 \blacksquare}$.

$\color{gray}{\text{Proof for PV}}$
$$\color{gray}{\begin{align}
\sum_{i=1}^n \frac 1{(1+r)^i}
&=\sum_{i=1}^n u^i\qquad\text{where $u=\frac 1{1+r}$}\\
&=\frac {1-u^n}{1-u}\\
&=\frac 1r\left[1-\frac 1{(1+r)^n}\right]\\
\text{PV}=\sum_{i=1}^n\frac C{(1+r)^i}&=\frac Cr\left[1-\frac 1{(1+r)^n}\right]\end{align}}$$


$\color{blue}{\text{Proof Using Recursion}}$
Let amount invested yearly $=C=2000$ and interest rate $=r=0.08$. 
From the recursion formula, 
$$\begin{align}
\require{cancel}
a_n&=(a_{n-1}+C)(\overbrace{1+r}^\lambda), \qquad a_0=0\\
&=\lambda a_{n-1}+\lambda C\qquad\text{where }\lambda=1+r\\
a_n+\mu&=\lambda(a_{n-1}+\mu)
\qquad\text{where }\mu=\frac {\lambda}{\lambda-1}C=\frac{(1+r)}rC\\
&=\lambda^2(a_{n-2}+\mu)\\
&=\lambda^3(a_{n-3}+\mu)\\
&\qquad\vdots\\
&=\lambda^n (\cancelto{0}{a_0}+\mu)\\
a_n&=\mu(\lambda^n-1)\\
&=\frac{C(1+r)}r[(1+r)^n-1]=\color{red}{31,291\qquad\blacksquare}
\end{align}$$
