Future Value Annuity.

1. Payments made at the end of a term.

In this situation, one invests money at the end of a term(a month, a quarter, a semester or a year). In most cases, you will find that the installments are made at the end of a month.

Let us now try a situation where one invests money, \(x\), at the end of a term, and the money grows at interest, \(i\), for a certain number of years. The timeline representing the situation is as follows:

The accumulated amount of this investment is:

$${x(\,1+i)\,^2+x(\,1+i)\,+x(\, 1+i)\,^0}$$ $${x(\,1+i)\,^2+x(\,1+i)\,+x}$$

$${x(\,1+i)\,^2+x(\,1+i)\,+x(\, 1+i)\,^0}$$ $${x(\,1+i)\,^2+x(\,1+i)\,+x}$$

\begin{gather} x(\,1+i)\,^2+x(\,1+i)\,\\ +x(\, 1+i)\,^0\\ = x(\,1+i)\,^2+x(\,1+i)\,\\ +x \end{gather}

Rearranging:

$${x+x(\,1+i)\,+x(\,1+i)\,^2}$$

This looks like a geometric series. We will then need to find the variables that are required for a geometric series, i.e., \(a\) and \(r\).

\begin{gather*} a=x\\ r=\frac{x(\,1+i)\,}{x}\\ =1+i, r>1 \end{gather*}

The sum of the geometric series formula that we will use is:

$${S_n= \frac{a(\,r^{n}-1)\,}{r-1}}$$

Now, substituting the variables obtained earlier into the above.

\begin{alignat*}{1} S_n= \frac{x\left[(1+i)^{n}-1\right]}{(1+i)-1}\\ S_n=\frac{x\left[(1+i)^{n}-1\right]}{i} \end{alignat*}

When we define the sum, \(S_n\), as the future value, \(F\), the above equation becomes:

$${F= \frac{x\left[(\,1+i)\,^{n}-1\right]}{i}}$$

where \(F\) is the future value, \(x\) is the monthly installment, \(i\) is the interest, and \(n\) is the number of installments.

Example: Let us see if the two methods used above can give us the same answer if we invest R1000 at the end of each year. How much money will we have at the end of three years if the bank gives us interest of 10% p.a?

The first method requires us to to note that \(i=10\% = 0.1\) and \(x=R1000\). The answer is:

\begin{gather} x+x(\,1+i)\,+x(\,1+i)\,^2\\ =1000+1000(\,1+0.1)\,+1000(\,1+0.1)\,^2\\ =R3310 \end{gather}

\begin{gather} x+x(\,1+i)\,+x(\,1+i)\,^2\\ =1000+1000(\,1+0.1)\,+1000(\,1+0.1)\,^2\\ =R3310 \end{gather}

\begin{gather} x+x(\,1+i)\,+x(\,1+i)\,^2\\ =1000+1000(\,1+0.1)\,\\ +1000(\,1+0.1)\,^2\\ =R3310 \end{gather}

Before using the formula, let us give you a tip on the \(n\), the number of installements.

$${n =last-first+1}$$

The number of installments are not always easy to find, especially for a large sample. So when we have drawn the appropriate timeline, we will be able to get them right. For our example:

$${n =3-1+1}$$

Now

\begin{gather} F=\frac{x\left[(\,1+i)\,^{n}-1\right]}{i}\\ =\frac{1000\left[(\,1+0.1)\,^{3}-1\right]}{0.1}\\ =R3310 \end{gather}

2. Payments made at the beginning of a term.

Using the parameters of the first case, the timeline for the second case is as follows

The accumulated amount of this investment is:

\begin{gather} x(\,1+i)\,^3+x(\,1+i)^2\,+x(\, 1+i)\, \end{gather}

\begin{gather} x(\,1+i)\,^3+x(\,1+i)^2\,+x(\, 1+i)\, \end{gather}

\begin{gather} x(\,1+i)\,^3+x(\,1+i)^2\,\\ +x(\, 1+i)\, \end{gather}

Rearranging:

$${x(\,1+i)\,+x(\,1+i)\,^2+x(\,1+i)\,^3}$$

$${x(\,1+i)\,+x(\,1+i)\,^2+x(\,1+i)\,^3}$$

\begin{gather} x(\,1+i)\,+x(\,1+i)\,^2\\+x(\,1+i)\,^3 \end{gather}

Now,

\begin{gather*} a=x(\,1+i)\,\\ r=\frac{x(\,1+i)\,^2}{x(\,1+i)\,}\\ =1+i, r>1 \end{gather*}

Substituting into the sum of geometric series formula:

\begin{gather} S_n= \frac{a(\,r^{n}-1)\,}{r-1}\\ F= \frac{x(1+i)\left[(1+i)^{n}-1\right]}{(1+i)-1}\\ F=\frac{x\left[(1+i)^{n}-1\right]}{i}(1+i) \end{gather}

Now let us check the solutions.

\begin{gather} x(\,1+i)\,+x(\,1+i)\,^2+x(\,1+i)\,^3\\ =1000(\,1+0.1)\,+1000(\,1+0.1)\,^2+1000(\,1+0.1)\,^3\\ =R3641 \end{gather}

\begin{gather} x(\,1+i)\,+x(\,1+i)\,^2+x(\,1+i)\,^3\\ =1000(\,1+0.1)\,+1000(\,1+0.1)\,^2+1000(\,1+0.1)\,^3\\ =R3641 \end{gather}

\begin{gather} x(\,1+i)\,+x(\,1+i)\,^2+x(\,1+i)\,^3\\ =1000(\,1+0.1)\,+1000(\,1+0.1)\,^2\\ +1000(\,1+0.1)\,^3\\ =R3641 \end{gather}

Using the formula:

$${n=2-0+1=3}$$ \begin{gather} F=\frac{x\left[(1+i)^{n}-1\right]}{i}(1+i)\\ =\frac{1000\left[(1+0.1)^{3}-1\right]}{0.1}(1+0.1)\\ =R3641 \end{gather}

3. Payment made immediately and thereafter made at the end of each term

The accumulated amount of this investment is:

\begin{gather} x(\,1+i)\,^3+x(\,1+i)^2\,+x(\, 1+i)\,^1+x(\, 1+i)^0\,\\ =x(\,1+i)\,^3+x(\,1+i)^2\,+x(\, 1+i)^1+x \end{gather}

\begin{gather} x(\,1+i)\,^3+x(\,1+i)\,^2 +x(\, 1+i)^1+x(\, 1+i)\,^0\\ =x(\,1+i)\,^3+x(\,1+i)\,^2+x(\, 1+i)\,^1+x \end{gather}

\begin{gather} x(\,1+i)\,^3+x(\,1+i)\,^2 \\+x(\, 1+i)\,^1+x(\, 1+i)\,^0\\ =x(\,1+i)\,^3+x(\,1+i)\,^2\\ +x(\, 1+i)\,+x \end{gather}

For the geometric series formula, and having rearranged the above:

\begin{gather*} a=x\\ r=\frac{x(\,1+i)\,^2}{x(\,1+i)\,}\\ =1+i, r>1 \end{gather*}

The sum of geometric series formula

\begin{gather} S_n= \frac{a(\,r^{n}-1)\,}{r-1}\\ F= \frac{x\left[(1+i)^{n}-1\right]}{(1+i)-1}\\ F=\frac{x\left[(1+i)^{n}-1\right]}{i} \end{gather}

Checking the solutions:

\begin{gather} =x+x(\,1+i)\,+x(\,1+i)\,^2+x(\, 1+i)^3\\ =1000+1000(\,1+0.1)\,+1000(\,1+0.1)\,^2+1000(\,1+0.1)\,^3\\ =R4641 \end{gather}

\begin{gather} =x+x(\,1+i)\,+x(\,1+i)\,^2+x(\, 1+i)\,^3\\ =1000+1000(\,1+0.1)\,+1000(\,1+0.1)\,^2\\ +1000(\,1+0.1)\,^3\\ =R4641 \end{gather}

\begin{gather} =x+x(\,1+i)\,+x(\,1+i)\,^2\\ +x(\, 1+i)^3\\ =1000+1000(\,1+0.1)\,+1000(\,1+0.1)\,^2\\ +1000(\,1+0.1)\,^3\\ =R4641 \end{gather}

Using the formula:

$${n=3-0+1=4}$$ \begin{gather} F=\frac{x\left[(1+i)^{n}-1\right]}{i}(1+i)\\ =\frac{1000\left[(1+0.1)^{4}-1\right]}{0.1}(1+0.1)\\ =R4641 \end{gather}

NB: Others prefer using the following equation when payment is made immediately:

$${F=\frac{x\left[(1+i)^{n+1}-1\right]}{i}}$$

The \(n\) here is calculated as the normal number of years multiplied by the rate at which the interest is quoted.

4. Payments made at the end of a term but missed the last payment.

The accumulated amount of this investment is:

$${x(\,1+i)\,^2+x(\,1+i)\,^1}$$

Rearranging:

$${x(\, 1+i)\,+x(\,1+i)\,^2}$$

For the sum of the geometric series formula::

\begin{gather*} a=x(\, 1+i)\,\\ r=\frac{x(\,1+i)\,^2}{x(\,1+i)\,}\\ =1+i, r>1 \end{gather*}

Then

\begin{gather} S_n= \frac{a(\,r^{n}-1)\,}{r-1}\\ F= \frac{x(\, 1+i)\,\left[(1+i)^{n}-1\right]}{(1+i)-1}\\ F=\frac{x\left[(1+i)^{n}-1\right]}{i}(\, 1+i)\, \end{gather}

Now let us check the solutions

\begin{gather} x(\, 1+i)\,+x(\,1+i)\,^2\\ =1000(\,1+0.1)\,+1000(\,1+0.1)\,^2\\ =R2310 \end{gather}

Using the formula:

$${n=2-1+1=2}$$ \begin{gather} F=\frac{x\left[(1+i)^{n}-1\right]}{i}(1+i)\\ =\frac{1000\left[(1+0.1)^{2}-1\right]}{0.1}(1+0.1)\\ =R2310 \end{gather}