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In actuarial mathematics, one uses the symbol \(a_{\overline{n}|i}\) to refer to the present value of an annuity-immediate with \(n\) periods, payment of 1 per period, and an effective interest rate of \(i\) per period. The formula for \(a_{\overline{n}|i}\) is:
\[a_{\overline{n}|i} = \frac{1 – v^n}{i}\]
I will derive that formula here. My derivation will not be efficient, because I want it to be as transparent as possible even to new learners who might be rusty with their algebra.
Using another standard notation, let \(v = \frac{1}{1+i}\), where \(i\) is the effective interest rate per period. The present value of the mth payment of our annuity is equal to \(v^m\): you can think of \(v^m\) as what you would get if you were to pull the payment (of 1) back through time to time 0, discounting it once per period along the way. The process is the reverse of what happens with compound interest.
Since an annuity consists of its payments, the present value of an annuity is the sum of the present values of each of its payments. In our case, that’s
\[v^1 + v^2 + … + v^n\]
This is a finite geometric series. Let’s divide through by v:
\[v (1 + v + … + v^{n-1}) \]
Now we apply the formula for the sum of a basic finite geometric series to get
\[v \frac{1 -v^n}{1-v}\]
We can see \( 1 – v^n \) in the numerator, so we’re already halfway to our formula. What remains is to convert the remaining terms into \(\frac{1}{i}\). Since there is no explicit \(i\) in the starting formula, let’s write out \(v\) as \(\frac{1}{1+i}\) everywhere except in the term we want to keep:
\[\color{green}{\frac{i}{1+i}}\cdot\frac{1- v^n}{1-\color{green}{\frac{1}{1+i}}}\]
The messiness of the denominator may hinder insight, so we will want to clean it up by combining terms. This requires us first to write out the 1 as \(\frac{1+i}{1+i}\):
\[\frac{1}{1+i}\cdot\frac{1- v^n}{\color{green}{\frac{1+i}{1+i}}-\frac{1}{1+i}}\]
And now combine terms:
\[\frac{1}{1+i}\cdot\frac{1- v^n}{\color{green}{\frac{1+i-1}{1+i}}}\]
Do some very basic arithmetic, namely \(1 – 1\) (a computation so simple that not even I can screw it up) to get:
\[\frac{1}{1+i}\cdot\frac{1- v^n}{\frac{\color{green}{i}}{1+i}}\]
And then multiply straight through to get:
\[\frac{1- v^n}{i}\]
Which is what we were after:
\(a_{\overline{n}|i} = \frac{1 – v^n}{i}\)