Question

Finding Account Balances In Exercise,complete the table to determine the table A for P dollars invested at rate r for t years,compounded n times per year .see Example 3.
n 1 2 4 12 365 continuous compounding
A
p = $1000,r = 4%,t = 20 years

Answer 1

Answer:

n = 1, A = $2,191.12

n = 2, A = $2,208.04

n = 4, A = $2216.71

n = 12, A = $2222.58

n = 365, A = $2225.44

Compound continuously, A = $2,225.54

Step-by-step explanation:

We are given the following in the question:

P = $1000

r = 4% = 0.04

t = 20 years

Formula:

The compound interest is given by

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

where P is the principal, r is the interest rate, t is the time, n is the nature of compound interest and A is the final amount.

For n = 1

$A = 1000\bigg(1 + \displaystyle\frac{0.04}{1}\bigg)^{20}\\\\A = \ 2,191.12$

For n = 2

$A = 1000\bigg(1 + \displaystyle\frac{0.04}{2}\bigg)^{40}\\\\A = \ 2,208.04$

For n = 4

$A = 1000\bigg(1 + \displaystyle\frac{0.04}{4}\bigg)^{80}\\\\A = \ 2216.71$

For n = 12

$A = 1000\bigg(1 + \displaystyle\frac{0.04}{12}\bigg)^{240}\\\\A = \ 2222.58$

For n = 365

$A = 1000\bigg(1 + \displaystyle\frac{0.04}{365}\bigg)^{7300}\\\\A = \ 2225.44$

Compounded continuously:

$A = Pe^{rt}\\A = 1000e^{0.04\times 20}\\A = 2,225.54$