Algebra 2 - compounding

Question

Seth's parents give him $5,000 to invest for his 16th birthday. He is considering two investment options: Option A will pay him 4.5% interest compounded continuously, and Option B will pay him 4.6% compounded quarterly. Write a function with Option A and Option B that calculates the value of each account after N years.

Answer 1

Hi Lillie,

So for compounding, we have the following formulas:

Continuous compounding: A = Pe^rt where A = amount, P = principal, r = interest rate, t = time in years, e = constant e

Normal compounding: A = P(1 + r/n)^nt where A = amount, P = principal, r = interest rate, t = time in years, n = number of times interest if compounded per year

For option A, we would need to use the continuous compounding formula. Plugging in, we have:

P = 5000, e = 2.71828, r = 4.5% = 0.045, t = N years

A = Pe^rt = 5000(2.71828)^0.045N. That's the formula, or if you do not want to plug in for e, A = 5000(e)^0.045N.

For option B, we use the normal compounding formula. Plugging in, we have:

P = 5000, r = 4.6% = 0.046, t = N years, n = 4 since there are 4 quarters in a year.

A = P(1 + r/n)^nt = 5000(1+ 0.046/4)^4t. That's the formula for option B.

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Answer 2

Plan A

A = 5000e^.045N for interest compounded continuosly with initial investment $5000 for N years at 4.5% annual interest

Plan B

B=5000(1+ r/n)^nN where n=4= number of compounding periods per year, r= annual interest rate= 4.6%, M=Number of years

B=5000(1+.046/4)^4M

=5000(1.0115)^4M

they both increase the investment to roughly the same amount

the more compounding periods per year, the closer to continuously compounding

5000(e^.045N) = 5000(1.0115)^4M

e^.045N = 1.0115^4M

.045N = 4Mln1.0115

N= M(4ln1.0115)/.045

N= 1.0164M when both Plans have the same end resulting return on investment