Question

Suppose you make a deposit of $P into a savings account that earns interest at a rate of 100r % per year. Show that if interest is compounded once per year, then the balance after t years is B(t) = P(1 + r)ᵗ. If interest is compounded m times per year, then the balance after t years is B(t) = P(1 + r/m)ᵐᵗ. For example, m = 12 corresponds to monthly compounding, and the interest rate for each month is r/12. In the limit m rightarrow infinity, the compounding is said to be continuous. Show that with continuous compounding, the balance after t years is B(t) = Peʳᵗ

Answer 1

The balance of a deposit after t years with annual compounding interest is calculated with B(t) = P(1 + r)^t, with more frequent compounding represented by B(t) = P(1 + r/m)^mt, and for continuous compounding by B(t) = Pe^rt.

Step-by-step explanation:

If you make a deposit of $P into a savings account that earns interest at a rate of 100r% per year, the balance after t years with interest compounded annually can be calculated using the formula B(t) = P(1 + r)t. This illustrates that the interest is added to the initial deposit, and this total becomes the new principal for the next period.

For interest compounded m times per year, the formula becomes B(t) = P(1 + r/m)mt, where m is the frequency of compounding per year, and r is the annual interest rate divided by m. For example, if m = 12, it corresponds to monthly compounding, and each month the interest rate applied is r/12.

When m approaches infinity, the compounding is considered to be continuous, and the balance after t years is given by the limit of the compound interest formula, yielding B(t) = Pert, where e is the base of the natural logarithm (approximately 2.71828).