different to the above worked solution.
Example 2: Ben wants to buy a new car. Rather than take out a loan, he decides to save up the money and invests $\$ 200$ at the end of each month in an account earning $3 \%$ interest per monthly. How much will he have saved up after 2 years?
i) Use a geometric series to answer this question.
ii) Verify your solution through the use of the Future Value of an Annuity Table in Appendix A.

Step 1 ：Ben invests $200 at the end of each month for 2 years, so there are a total of 24 payments.

Step 2 ：The interest rate is 3% per month, so the monthly interest rate is $\frac{3}{100}=0.03$.

Step 3 ：We can use the formula for the sum of a geometric series to calculate the total amount of money after 2 years. The formula is $S = a \cdot \frac{1 - r^n}{1 - r}$, where $a$ is the first term, $r$ is the common ratio, and $n$ is the number of terms.

Step 4 ：In this case, the first term $a$ is the amount of the first investment, which is $200$. The common ratio $r$ is $1 + 0.03 = 1.03$, and the number of terms $n$ is the number of payments, which is 24.

Step 5 ：Substitute these values into the formula, we get $S = 200 \cdot \frac{1 - (1.03)^{24}}{1 - 1.03}$.

Step 6 ：Solving the above equation, we get $S = 200 \cdot \frac{1 - (1.03)^{24}}{-0.03}$.

Step 7 ：Solving the above equation, we get $S = -200 \cdot (1 - (1.03)^{24}) \div 0.03$.

Step 8 ：Solving the above equation, we get $S = -200 \cdot (1 - 10.243) \div 0.03$.

Step 9 ：Solving the above equation, we get $S = -200 \cdot (-9.243) \div 0.03$.

Step 10 ：Solving the above equation, we get $S = 200 \cdot 9.243 \div 0.03$.

Step 11 ：Solving the above equation, we get $S = 1848.6 \div 0.03$.

Step 12 ：Solving the above equation, we get $S = \boxed{61620}$.

Step 13 ：To verify this solution, we can use the Future Value of an Annuity Table in Appendix A. Look up the future value factor for 24 periods at 3% interest rate, and multiply it by the monthly investment of $200$. The result should be the same as the one we calculated.