# Future Value of Ordinary Annuity equation

I'm learning precalculus via an open source book and it states something I'm confused about. Here is the quote:

Suppose you have an account with annual interest rate r which is compounded n times per year. We let i = $$\frac{r}{n}$$ denote the interest rate per period. Suppose we wish to make ongoing deposits of P dollars at the end of each compounding period. Let $$A_k$$ denote the amount in the account after k compounding periods. Then $$A_1$$ = P, because we have made our first deposit at the end of the first compounding period and no interest has been earned. During the second compounding period, we earn interest on $$A_1$$ so that our initial investment has grown to $$A_1(1 + i)$$ = P(1 + i) in accordance with Equation 6.1. When we add our second payment at the end of the second period, we get $$A_2 = A_1(1 + i) + P = P(1 + i) + P = P(1 + i)(1 + \frac{1}{1+i})$$.

It then goes on showing iterations of $$A_3, A_4...$$

Now, I get that at $$A_2$$ we have $$A_1(1 + i) + P$$. And I get this is equal to P(1 + i) + P even more so because P(1 + i) is our initial principal plus one compounding period of interest and the next + P is the additional principal we added this time around.
What I don't get is how does P(1 + i) + P = P(1 + i)(1 + $$\frac{1}{1+i})$$?

If I try plugging numbers into this it doesn't add up. Lets say P = 10 and lets say i = 2%

then P(1 + i) + P = 10(1 + .2) + 10 = 20.20.

Now lets plug the same values into P(1 + i)(1 + $$\frac{1}{1+i})$$ =

10(1 + .2)(1 + $$\frac{1}{1+.2})$$ =

10.20(1 + $$\frac{1}{1+.2})$$ =

(10.20 + $$\frac{10.20}{1.2})$$ =

10.20 + 8.5 = 18.70.

And since these don't add up, I don't see how P(1 + i) + P = P(1 + i)(1 + $$\frac{1}{1+i})$$? Am I doing wrong here?

• To begin with $10(1.2)=12$, not $10.2$ – saulspatz Jan 22 '20 at 22:40
• to add to the list 2% is 0.02 not 0.2 – user645636 Jan 22 '20 at 22:44

The expression you are querying has simply used $$(1+i) \frac{1}{1+i} = 1$$