Q. 61 - Calculus

Question

Suppose you invest $100.00 in a bank that pays you a nominal annual interest rate of 6%. The bank offers you the option of compounding your money n times over the course of a year. Your balance in the account after one year is given by

$a_k = 100.00{\left(1 + \frac{0.06}{k}\right)}^k$

Find the balance in the account, rounded to the nearest cent, if you compound the interest k times during the year, where k is 1, 2, 6, 12, 52, and 365. You should see that the balance at the end of the year is increasing. Do you believe the sequence is bounded? If so, by what

value? Be as specific as possible.

Step-by-Step Solution

Answer

If k=1, $a_k=106$

If k=2, $a_k=106.09$

If k=6, $a_k=106.15$

If k=12, $a_k=106.17$

If k=52,$a_k=106.18$

If k=365,$a_k=106.18$

Step 1. Given information

$a_k = 100.00{\left(1 + \frac{0.06}{k}\right)}^k$

Step 2. Calculate for k=1

$a_k = 100.00{\left(1 + \frac{0.06}{k}\right)}^k=100.00{\left(1 + \frac{0.06}{1}\right)}^1$

$a_k = 100.00{\left(1 + \frac{0.06}{k}\right)}^k=106.06$

Calculate for k=2

$a_k = 100.00{\left(1 + \frac{0.06}{k}\right)}^k=100.00{\left(1 + \frac{0.06}{2}\right)}^2$

$a_k = 100.00{\left(1 + \frac{0.06}{k}\right)}^k=106.15$

Calculate k=6

$a_k = 100.00{\left(1 + \frac{0.06}{k}\right)}^k= 100.00{\left(1 + \frac{0.06}{6}\right)}^6$

$a_k = 100.00{\left(1 + \frac{0.06}{k}\right)}^k=106.17$

Step 5. Calculate for k=12

$a_k = 100.00{\left(1 + \frac{0.06}{k}\right)}^k=100.00{\left(1 + \frac{0.06}{12}\right)}^{12}$

$a_k = 100.00{\left(1 + \frac{0.06}{k}\right)}^k=106.18$

Step 6. Calculate for k=365

$a_k = 100.00{\left(1 + \frac{0.06}{k}\right)}^k=a 100.00{\left(1 + \frac{0.06}{365}\right)}^{365}$

$a_k = 100.00{\left(1 + \frac{0.06}{k}\right)}^k=106.18$