Question

Finding Account Balances In Exercise, complete the table to determine the balance A for P dollers invested at rate for years, compounded times per year.
n 1 2 4 12 365 Continuous compounding
A
P = $3000, r = 3.5%, t = 10 years

Answer 1

Answer:

For n=1  $A=4231.79$

For n=2  $A=4244.33$

For n=4  $A=4250.72$

For n=12 $A=4255.03$

For n=365 $A=4257.13$

Using exponential function $A=4257.20$

Step-by-step explanation:

P=$3000

r = 3.5%

t = 10 years

we need to find accumulated Amount A for n= 1, 2, 4, 12 and 365

As we know,  continuous compounding can be found by

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

Where P is the invested amount with interest rate r for t years

$A=3000(1+\frac{0.035}{n} )^{n*10}$

For n=1

$A=3000(1+\frac{0.035}{1} )^{1*10}=3000(1.410598) =4231.79$

For n=2

$A=3000(1+\frac{0.035}{2} )^{2*10}=3000(1.414778) =4244.33$

For n=4

$A=3000(1+\frac{0.035}{4} )^{4*10}=3000(1.416908) =4250.72$

For n=12

$A=3000(1+\frac{0.035}{12} )^{12*10}=3000(1.418344) =4255.03$

For n=365

$A=3000(1+\frac{0.035}{365} )^{365*10}=3000(1.419043) =4257.13$

As the value of n increases, the change in the value of A decreases and eventually for very large value of n, A becomes constant and doesnt change anymore.

This is why we also use exponential function for continuous compounding  

$A=Pe^{rt}$

$A=3000e^{0.035*10}=3000(1.41906)=4257.20$

Hence we got same result with above equation and proved that exponential function provides accurate results for very large value of n.