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

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

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(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(0.04)/(1)\bigg)^(20)\\\\A = \$2,191.12`

For n = 2

`A = 1000\bigg(1 + \displaystyle(0.04)/(2)\bigg)^(40)\\\\A = \$2,208.04`

For n = 4

`A = 1000\bigg(1 + \displaystyle(0.04)/(4)\bigg)^(80)\\\\A = \$2216.71`

For n = 12

`A = 1000\bigg(1 + \displaystyle(0.04)/(12)\bigg)^(240)\\\\A = \$2222.58`

For n = 365

`A = 1000\bigg(1 + \displaystyle(0.04)/(365)\bigg)^(7300)\\\\A = \$2225.44`

Compounded continuously:

`A = Pe^(rt)\\A = 1000e^(0.04* 20)\\A = $2,225.54`