Question

Finding Account Balances In Exercise, complete the table to determine the balance A for P dollars Invested at rate for years compounded time per year.

n 1 2 4 12 365 Continuous compounding
A
P = $1000, r = 4%, t = 5 years

Answer 1

n = 1, A = $1,216.66

n = 2, A = $1,218.99

n = 4, A = $1,220.1

n = 12, A = $1,221.00

n = 365, A = $1,221.39

Compounded continuously, A = $1,221.40

Explanation:

We are given the following in the question:

P = $1000

r = 4% = 0.04

t = 5 years

The compound interest is given by:

`A = p\bigg(1+(r)/(n)\bigg)^(nt)`

where A is the amount, p is the principal, r is the interest rate, t is the time in years and n is the nature of compound interest.

When compounded continuously:

`A = pe^(rt)`

where A is the amount, p is the principal, r is the interest rate, t is the time in years

For n = 1

`A = 1000\bigg(1+(0.04)/(1)\bigg)^5\\\\A = \$1,216.66`

For n = 2

`A = 1000\bigg(1+(0.04)/(2)\bigg)^(10)\\\\A = \$1,218.99`

For n = 4

`A = 1000\bigg(1+(0.04)/(4)\bigg)^(20)\\\\A = \$1,220.19`

For n = 12

`A = 1000\bigg(1+(0.04)/(12)\bigg)^(60)\\\\A = \$1,221.00`

For n = 365

`A = 1000\bigg(1+(0.04)/(365)\bigg)^(1825)\\\\A = \$1,221.39`

Compounded continuously

`A = 1000e^(5* 0.04)\\A = \$1,221.40`