**Answer:**

\begin{array}{|c|c|}\cline{1-2} \vphantom{\dfrac12} n&A

\cline{1-2}\vphantom{\dfrac12} 1& \$4376.10

\vphantom{\dfrac12} 2& \$4442.10

\vphantom{\dfrac12} 4& \$4476.86

\vphantom{\dfrac12} 12& \$4500.73

\vphantom{\dfrac12} 365& \$4512.49

\vphantom{\dfrac12} \sf Continuous& \$4512.89

\cline{1-2} \end{array}

**Step-by-step explanation:**

\boxed{\begin{minipage}{8.5 cm}\underline{Compound Interest Formula}

$ A=P\left(1+\frac{r}{n}\right)^{nt}$

where:

\phantom{ww}$\bullet$ $A =$ final amount

\phantom{ww}$\bullet$ $P =$ principal amount

\phantom{ww}$\bullet$ $r =$ interest rate (in decimal form)

\phantom{ww}$\bullet$ $n =$ number of times interest is applied per year

\phantom{ww}$\bullet$ $t =$ time (in years)

\end{minipage}}

Given:

- P = $2100
- r = 8.5% = 0.085
- t = 9 years

**Substitute **the given **values **into the **formula **to create an **equation for A **in terms of n:

\implies A=2100\left(1+\dfrac{0.085}{n}\right)^{9n}

**Substitute **each **value of n** into the **equation**:

\begin{aligned}n=1 \implies A&=2100\left(1+\dfrac{0.085}{1}\right)^{9 \times 1}

&=2100\left(1.085\right)^{9}

&=\$4376.10\end{aligned}

\begin{aligned}n=2 \implies A&=2100\left(1+\dfrac{0.085}{2}\right)^{9 \times 2}

&=2100\left(1.0425\right)^{18}

&=\$4442.10\end{aligned}

\begin{aligned}n=4 \implies A&=2100\left(1+\dfrac{0.085}{4}\right)^{9 \times 4}

&=2100\left(1.02125\right)^{36}

&=\$4476.86\end{aligned}

\begin{aligned}n=12 \implies A&=2100\left(1+\dfrac{0.085}{12}\right)^{9 \times 12}

&=2100\left(1.00708333...\right)^{108}

&=\$4500.73\end{aligned}

\begin{aligned}n=365 \implies A&=2100\left(1+\dfrac{0.085}{365}\right)^{9 \times 365}

&=2100\left(1.00023287...\right)^{3285}

&=\$4512.49\end{aligned}

\boxed{\begin{minipage}{8.5 cm}\underline{Continuous Compounding Formula}

$ A=Pe^{rt}$

where:

\phantom{ww}$\bullet$ $A =$ final amount

\phantom{ww}$\bullet$ $P =$ principal amount

\phantom{ww}$\bullet$ $e =$ Euler's number (constant)

\phantom{ww}$\bullet$ $r =$ annual interest rate (in decimal form)

\phantom{ww}$\bullet$ $t =$ time (in years)

\end{minipage}}

\implies A=2100e^{0.085 \times 9}

\implies A=2100e^{0.765}

\implies A=2100(2.14899437...)

\implies A=\$4512.89

Input the **calculated values** into the **table**:

\begin{array}{|c|c|}\cline{1-2} \vphantom{\dfrac12} n&A

\cline{1-2}\vphantom{\dfrac12} 1& \$4376.10

\vphantom{\dfrac12} 2& \$4442.10

\vphantom{\dfrac12} 4& \$4476.86

\vphantom{\dfrac12} 12& \$4500.73

\vphantom{\dfrac12} 365& \$4512.49

\vphantom{\dfrac12} \sf Continuous& \$4512.89

\cline{1-2} \end{array}