Ordinary Annuity Future Value Formula Explained I am trying to conceptually understand the formula for the future value for an ordinary annuity.
I do not want to just plug and chug, but rather to understand why the formula $$\text{FV }=\text{ pymt }\cdot \frac{(1+i)^n - 1}{i}$$  works the way it does.

*

*Why do we add $1$ to $i$ in the numerator?

*Why do we raise this value to $n\,$?

*Why is $1$ then subtracted?

*Why is the total numerator divided by $i\,$?

 A: The formula arises from the sum of a finite geometric series.
Consider the accumulated value of a series of level payments of $K$ per period, in which the interest rate is $i$ per period.  Interest is compounded per period.  We want the accumulated value at the time of the $n^{\rm th}$ payment.
We work backwards:  the $n^{\rm th}$ payment has had $0$ periods to accrue interest because no time has elapsed between the valuation time and the time of payment.  Thus, its accumulated value is simply $K$.
The $(n-1)^{\rm th}$ payment has has $1$ period to accrue interest at rate $i$, thus its future value is the payment amount $K$ plus the interest accrued in one period, $Ki$.  So its future value is $K + Ki = K(1+i)$.
The $(n-2)^{\rm th}$ payment has had $2$ periods to accrue interest at rate $i$, and because interest is compounded per period, its accumulated value is determined by first finding its accumulated value after one period, then taking that as the amount to compound for the second period.  After one period, it has value $K(1+i)$ as we had shown above; after two periods, it has value $K(1+i)^2$, since by replacing $K$ with $K(1+i)$ in the previous period yields $K(1+i)^2$.
By continuing to extend this logic, we can reason that the total accumulated value of all level payments is the cash flow
$$K(1+i)^{n-1} + K(1+i)^{n-2} + \cdots + K(1+i)^2 + K(1+i) + K.$$
The reason why the first term is $K(1+i)^{n-1}$ is because this is the accumulated value of the first payment, which has had $n-1$ periods to accrue interest.
To simplify this sum, consider the finite geometric series  $$S_n(r) = 1 + r + r^2 + \cdots + r^{n-1}.$$  So long as $r \ne 1$, we may multiply both sides by $r-1$ to obtain
$$\begin{align*}
(r-1)S_n(r) &= (r-1)(1 + r + r^2 + \cdots + r^{n-1}) \\
&= (r + r^2 + r^3 + \cdots + r^n) - (1 + r + r^2 + \cdots + r^{n-1}) \\
&= r^n - 1. \end{align*}
$$
In the first step, we distributed the product over the terms in $r-1$.  In the second step, we observed that there is a cancellation of intermediate terms in the sum, except for $r^n$ and $-1$.
Therefore, for $r \ne 1$, we have the formula
$$S_n(r) = 1 + r + r^2 + \cdots + r^{n-1} = \frac{r^n - 1}{r - 1}.$$
Now substituting $r = 1 + i$, and multiplying both sides by $K$, we obtain the desired accumulated value formula
$$KS_n(1 + i) = K(1 + (1+i) + (1+i)^2 + \cdots + (1+i)^{n-1}) = K \frac{(1+i)^n - 1}{i}.$$
To test your understanding, note that the formula for $S_n(r)$ is invalid for $r = 1$, since we then have division by zero.  What interest rate corresponds to such an $r$?  What is the accumulated value of such a cash flow at such an interest rate?  Clearly, some accumulated value must exist, yet the formula does not work in such a case; what would it be?
