Sebastian deposits $\$ 1,300$ every quarter into an account earning an annual interest rate of $7.1 \%$ compounded quarterly. How much would he have in the account after 12 years, to the nearest dollar? Use the following formula to determine your answer.
\[
A=d\left(\frac{(1+i)^{n}-1}{i}\right)
\]
$A=$ the future value of the account after $n$ periods
$d=$ the amount invested at the end of each period
$i=$ the interest rate per period
$n=$ the number of periods

Step 1 ：Given that Sebastian deposits $1300 every quarter into an account earning an annual interest rate of 7.1% compounded quarterly, we are to find how much he would have in the account after 12 years, to the nearest dollar.

Step 2 ：We use the formula for the future value of a series of payments, also known as an annuity, which is given by \[A=d\left(\frac{(1+i)^{n}-1}{i}\right)\] where \(A\) is the future value of the account after \(n\) periods, \(d\) is the amount invested at the end of each period, \(i\) is the interest rate per period, and \(n\) is the number of periods.

Step 3 ：First, we calculate the interest rate per period, \(i\), by dividing the annual interest rate by the number of periods in a year. Since the interest is compounded quarterly, there are 4 periods in a year. Therefore, \(i = \frac{7.1}{100 \times 4} = 0.01775\).

Step 4 ：Next, we calculate the number of periods, \(n\), by multiplying the number of years by the number of periods in a year. Therefore, \(n = 12 \times 4 = 48\).

Step 5 ：Substituting \(d = 1300\), \(i = 0.01775\), and \(n = 48\) into the formula, we get \[A = 1300 \times \left(\frac{(1+0.01775)^{48}-1}{0.01775}\right)\].

Step 6 ：Solving the above expression, we get \(A = 97180\).

Step 7 ：Rounding to the nearest dollar, we get \(A = 97180\).

Step 8 ：Final Answer: Sebastian would have \(\boxed{97180}\) dollars in the account after 12 years, to the nearest dollar.