# Why do we take the no. of years to compound the interest as power?

1. Feb 1, 2010

### Juwane

Why do we take the "no. of years to compound the interest" as power?

Suppose interest is given at 12% annually, compounded once a year. At the end of the year we will have (A = starting amount):

$$A( 1 + 0.12 )$$

But if it is compounded twice a year, then at the end of the year we will have:

$$2A \left( 1 + \frac{0.12}{2} \right)$$

Why is the above wrong? Why it should be $$A \left( 1 + \frac{0.12}{2} \right)^2$$ instead of $$2A \left( 1 + \frac{0.12}{2} \right)?$$

Last edited: Feb 1, 2010
2. Feb 1, 2010

### jambaugh

Re: Why do we take the "no. of years to compound the interest" as power?

The interest you earned for the first half year is now money upon which you earn more interest for the second half of the year. Take for example 20% interest on 100 dollars compounded twice a year. In the first half-year you earn 10% of $100 so you have$100+$10=$110.
For the second half-year you earn 10% on the $110 so you have$110 + $11 =$121.

That's what "compounded" means. You earn interest on the interest, and then interest on the interest on the interest, and so on.

It helps to forget "years" and work only with "periods". Say you earn interest per period at a rate of $$r_p$$. For a given period if you start with an amount $$A$$ at the end of that period you have amount $$A + r_p A= A(1+r_p)$$.

Do this for $$k$$ periods and you have:
$$A=A_0(1+r_p)(1+r_p)\cdots(1+r_p) = A_0(1+r_p)^k$$

Understand that formula first. Each compounding period effects a multiplication by 1 plus the rate. Repeated multiplications are expressed by a power.

Now we rescale to years. Given an annual rate of r and n periods in a year and t years.
$$r_p = r/n$$
$$k = nt$$
so
$$A(t) =A_0\left(1+r_p\right)^k= A_0 \left(1 + \frac{r}{n}\right)^{nt}$$

(with $$A_0 = P$$ the initial amount is called the "principle" by accountants.)