If you invest a dollar at “$\mathbf{\text{6\%}}$interest compounded monthly,” it amounts to ${\left(1.005\right)}^{\mathbf{n}}$dollars after months. If you invest $\mathbf{\$}\mathbf{10}$at the beginning of each month for 10 years (120 months), how much will you have at the end of the 10 years?

The monthly investment is ten dollars.

The amount at the end of n months, if 1 dollar is compounded at$6\%$ monthly is ${\left(1.005\right)}^n$.

A total of ten years has been put into this project.

A geometric series is a collection of numbers with a common ratio. This means that each phrase is found by multiplying the term before it.

The formula is provided by,

$\text{G.P}=\frac{a\left(r^n-1\right)}{r-1}$

Here,

a is the first term of series,

r is the common ratio,

n is the total time period.

When the investment is compounded monthly, the total amount is the sum of a geometric series of the form.

$a,ar,a{r}^2,....,a{r}^n$

The sum of G.P is provided by the expression,

$\text{G.P}=\frac{a\left(r^n-1\right)}{r-1}-----\left(1\right)$

If a person invests $\$1$at the beginning of each month for n months, he or she will have amassed a sum of money by the end of those months ${\left(1.005\right)}^n$. If you invest $\$10$at the beginning of the month, at the end of the first month, you will get $\$10\left(1.005\right)=\$10.05$.

A $\$10$investment is made for a ten-year period. As a result, the number of months the money is invested in is established.

$\begin{array}{rcl}n& =& \left(10\text{year}\right)\left(\frac{12\text{month}}{1\text{year}}\right)\\ & =& 120\text{months}\end{array}$

The amount received at the end of 120 months can be expressed using the geometric series below:

$S_n=\$10\left(1.005\right)+\$10{\left(1.005\right)}^2+\$10{\left(1.005\right)}^3+....+\$10{\left(1.005\right)}^{120}$

Comparing with the above series,

$S_n=a+ar+a{r}^2+....+a{r}^n$

Here,

$$\begin{gathered} a=10.05 \\ r=1.005 \\ n=120 \end{gathered}$$

Put the values of a, r and n in equation (1),

$\begin{array}{rcl}{S}_n& =& \frac{a\left(1-r^n\right)}{1-r}a\\ & =& \frac{\$\left(10.05\right)\left[1-{\left(1.005\right)}^{120}\right]}{1-\left(1.005\right)}\\ & =& \$1646.99\\ & & \end{array}$