Explicit formula for recursive geometric/arithmatic series

Question

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.

Answer 1

$\color{blue}{\text{Proof Using Discounting and Compounding}}$

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, $$\begin{align} \require{cancel} a_n&=(a_{n-1}+C)(\overbrace{1+r}^\lambda), \qquad a_0=0\\ &=\lambda a_{n-1}+\lambda C\qquad\text{where }\lambda=1+r\\ a_n+\mu&=\lambda(a_{n-1}+\mu) \qquad\text{where }\mu=\frac {\lambda}{\lambda-1}C=\frac{(1+r)}rC\\ &=\lambda^2(a_{n-2}+\mu)\\ &=\lambda^3(a_{n-3}+\mu)\\ &\qquad\vdots\\ &=\lambda^n (\cancelto{0}{a_0}+\mu)\\ a_n&=\mu(\lambda^n-1)\\ &=\frac{C(1+r)}r[(1+r)^n-1]=\color{red}{31,291\qquad\blacksquare} \end{align}$$

Answer 2

You seem to be trying to figure out how the total ammount of money in the bank account evolve over time, by looking at how one investment plus one year of interest changes that amount. This gives you a recursive formula that is outside of your curriculum to handle.

Now instead of looking at what happens to the amount of money in that single account, try to see this as $10$ completely separate investments into $10$ separate accounts. Then the first account will have $\$2000\cdot1.08^{10}$ in it. The second investment account has $\$2000\cdot 1.08^9$ in it, and so on, leading to the full expression $$ a_n = \sum_{i = 1}^n\$2000\cdot 1.08^i = \$2000\left(\sum_{i = 1}^n 1.08^i\right)\\ = \$2000 \cdot 1.08 \left(\sum_{i = 0}^{n-1}1.08^i\right) = \$2160\cdot \left(\sum_{i = 0}^{n-1}1.08^i\right) $$ Now, the last term here is the sum of a geometric series, and we know what that is: $$ \sum_{i = 0}^{n-1} 1.08^i = \frac{1-1.08^n}{1-1.08} $$ and for $n = 10$, we get that this is equal to $14.5$. Therefore, the total ammount of money in the account after $10$ years (moments before the eleventh investment) is $$ a_{10} = \$2160 \cdot 14.5 = \$31,\!291 $$

Answer 3

Your formula is correct. More general: If $a_0$ is the constant investment every year (in your case $a_0=2000$), and if $p$ is the multiplicator of the annual interest (here $p=1.08$), then you have $a_{n}=(a_0+a_{n-1})\cdot p$ for each $n\in\mathbb N$. Let's calulcate the first values of $a_n$: \begin{align*} a_1&=(a_0+a_0)p=2a_0 p \\ a_2&=(a_0+a_1)p=a_0p+2a_0p^2 \\ a_3&=(a_0+a_2)p=a_0p+(a_0p+2a_0p^2)p=a_0p+a_0p^2+2a_0p^3 \end{align*} and so on. With induction, one can prove that \begin{align*} a_n=\sum_{k=1}^na_0p^k=a_0\cdot\frac{p(p^n-1)}{p-1}, \end{align*} which should be the desired formula.

Answer 4

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 $C = 27000$.