We need to find the accumulated value A of an investment of P =$10000 for t = 6 years at an interest rate of r = 4% = 0.04.
When the investment is compounded n times a year, A is given by:
[tex]A=P\left(1+\frac{r}{n}\right)^{nt}[/tex]When it is compounded continuously, we need to use the limit of the above formula for n going to infinity. Then, A is given by:
[tex]A=Pe^{rt}[/tex]Part a
When the investment is compounded semiannually, n = 2. Then, we have:
[tex]A=10000\left(1+\frac{0.04}{2}\right)^{2\cdot6}=10000\left(1.02\right)^{12}\cong12682.42[/tex]Answer (in dollars): 12682.42
Part b
When the investment is compounded quarterly, n = 4. Then, we have:
[tex]A=10000\left(1+\frac{0.04}{4}\right)^{4\cdot6}=10000\left(1.01\right)^{24}\cong12697.35[/tex]Answer (in dollars): 12697.35
Part c
When the investment is compounded monthly, n = 12. Then, we have:
[tex]A=10000\left(1+\frac{0.04}{12}\right)^{12\cdot6}=10000\left(1+\frac{0.01}{3}\right)^{72}\cong12707.42[/tex]Answer (in dollars): 12707.42
Part d
When the investment is compounded continuously, we have:
[tex]A=10000(e^{0.04\cdot6})=10000(e^{0.24})=12712.49[/tex]Answer (in dollars): 12712.49
