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}