# Compound Interest Rate

## General Formula

Given an annual interest rate r, we can use the number following formula to compute the increase in the principle when the interest is compound annually.

${\displaystyle A(t)=P(1+r)^{t}}$

If the interest is compounded in n times a year (in our example n=12), the amount after t years is,

${\displaystyle A(t)=P({\frac {1+r}{n}})^{tn}}$

If we take a limit as n goes to infinity, we say that interest is compounded continuously.

As we know, ${\displaystyle e=\lim _{n\to \infty }(1+{\frac {1}{n}})^{n}}$

Let's use the substitution m=n/r, so n=mr

{\displaystyle {\begin{aligned}A(t)&=\lim _{m\to \infty }P(1+{\frac {1}{m}})^{tmr}\\&=P(\lim _{m\to \infty }(1+{\frac {1}{m}})^{tmr}\\&=P(\lim _{m\to \infty }((1+{\frac {1}{m}})^{m})^{tr}\\&=Pe^{tr}\end{aligned}}}

## Practice Question

Q: say P=\$10,000 with r=0.06 and t=3, how much will you have if interest is compounded.

(1) 6 times a year

(2) 24 times a year

(3) continuously

A:

(1) ${\displaystyle A=P(1+r/n)^{tn}=10000(1+0.06/6)^{3\times 6}=11961.47}$

(2) ${\displaystyle A=10000(1+0.06/24)^{3\times 24}=11969.48}$

(3) ${\displaystyle A=Pe^{rt}=10000e^{0.06\times 3}=11972.17}$