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

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`

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+(r)/(n) )^(nt)`

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

`A=3000(1+(0.035)/(n) )^(n*10)`

For n=1

`A=3000(1+(0.035)/(1) )^(1*10)=3000(1.410598) =4231.79`

For n=2

`A=3000(1+(0.035)/(2) )^(2*10)=3000(1.414778) =4244.33`

For n=4

`A=3000(1+(0.035)/(4) )^(4*10)=3000(1.416908) =4250.72`

For n=12

`A=3000(1+(0.035)/(12) )^(12*10)=3000(1.418344) =4255.03`

For n=365

`A=3000(1+(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.