To attend school, Chloe deposits $\mathrm{\$}400$ at the end of every quarter for two and one-half years. What is the accumulated value of the deposits if interest is $8\mathrm{\%}$ compounded annually?
The accumulated value is $\mathrm{\$}◻$.
(Round the final answer to the nearest cent as needed. Round all intermediate values to six decimal places as needed.)

Step 1 ：Calculate the interest rate per quarter by dividing the annual interest rate by 4: $\frac{0.08}{4}=0.02$

Step 2 ：Calculate the number of quarters in two and one-half years: $2.5\times 4=10$ quarters

Step 3 ：Calculate the accumulated value of each deposit. The first deposit accumulates interest for $10-1=9$ quarters, the second deposit accumulates interest for $10-2=8$ quarters, and so on. The last deposit does not accumulate any interest

Step 4 ：The accumulated value of each deposit is given by the formula ${(1+\frac{0.08}{4})}^n(\mathrm{\$}400)$, where n is the number of quarters the deposit accumulates interest

Step 5 ：Add up the accumulated value of all the deposits: ${(1+\frac{0.08}{4})}^9(\mathrm{\$}400)+{(1+\frac{0.08}{4})}^8(\mathrm{\$}400)+\dots +{(1+\frac{0.08}{4})}^1(\mathrm{\$}400)+\mathrm{\$}400$

Step 6 ：This is a geometric series with first term ${(1+\frac{0.08}{4})}^9(\mathrm{\$}400)$, common ratio $(1+\frac{0.08}{4})$, and 10 terms. The sum of a geometric series is given by the formula $\frac{a(1-r^n)}{1-r}$, where a is the first term, r is the common ratio, and n is the number of terms

Step 7 ：Substitute the values into the formula to get the final answer: $\frac{{(1+\frac{0.08}{4})}^9(\mathrm{\$}400)(1-{(1+\frac{0.08}{4})}^{10})}{1-(1+\frac{0.08}{4})}\approx \boxed{{\displaystyle \mathrm{\$}4,579.64}}$