In Algebra 1 you rewrote a yearly growth rate as a monthly one (Rewriting Exponential Expressions). Banks do exactly that when they compound interest several times a year.
Idea clave
$$A = P\left(1 + \frac{r}{n}\right)^{nt}$$ $P$ is the starting amount, $r$ the yearly rate as a decimal, $n$ the number of compounding periods per year and $t$ the number of years. Each period earns $\frac{r}{n}$, and there are $nt$ periods.
Por qué funciona
Interest earns interest. If a period's rate is $i$, the balance at the end of the period is the old balance plus $i$ times it, which is the old balance times $(1 + i)$. After $N$ periods you have multiplied by $(1 + i)$ $N$ times: $P(1 + i)^N$. With $n$ periods per year, $i = \frac{r}{n}$ and $N = nt$.
Worked examples
Ejemplo resuelto: Monthly compounding
\$2000 is invested at 4.8% per year, compounded monthly. Find the balance after 5 years.
- $r = 0.048$, $n = 12$, $t = 5$: rate per month $\frac{0.048}{12} = 0.004$, and $nt = 60$ months.
- $A = 2000(1.004)^{60}$ $\approx 2000(1.270641) \approx 2541.28$.
Respuesta: about \$2,541.28
Ejemplo resuelto: Comparing frequencies
\$1000 at 6% for 10 years: compare yearly and quarterly compounding.
- Yearly: $1000(1.06)^{10} \approx 1790.85$.
- Quarterly: $1000(1.015)^{40} \approx 1814.02$.
Respuesta: Quarterly gives about \$23 more.
Error común
Using the yearly rate for every period without dividing by $n$. Monthly compounding of 6% is $0.5\%$ per month, not $6\%$ per month. $1000(1.06)^{120}$ would claim over a million dollars after 10 years, which is clearly wrong. Divide the rate by $n$ and multiply the years by $n$.