## Fast Formulas #5: Quasi-Geometric Series (The Other Trick)

In Fast Formulas #3: Pool Average Life with CPR Prepayments,  I described Formula 3.1., a compact formula for the average life of a prepaying loan pool, and  I wrote that the formula had been found by using  “a couple of tricks.”    I explained one of the tricks in Fast Formulas #4: Geometric Series (One of the Tricks).  This post describes the other trick, which concerns a variation on the geometric series.

A geometric series is a sum such as $Q=q+q^{2}+q^{3}+...+q^{n-1}+q^{n}$

If n is large, it would be tedious to calculate the terms of $Q$.  Fortunately, there is a shortcut, as discussed in Fast Formulas #4: $Q=\dfrac {q-q^{n+1}}{1-q}$                                                                                                            (Formula 4.4)

If you multiply each term in the geometric series by its exponent, you get this variation: $S=q+2q^{2}+3q^{3}+...+(n-1)q^{n-1}+nq^{n}$

The series $S$ doesn’t have its own name, but it is an example of a quasi-geometric series.  It has applications to the duration of a bond or annuity.

We can find a shortcut formula for $S$ with the same strategy we used to find Formula 4.4. We construct a new series that shares most of its terms with the original series, and then cancel out the matching terms on our way to a compact formula.

The first step is to multiply $S$ by $q$: $qS=q^{2}+2q^{3}+3q^{4}+...+(n-1)q^{n}+nq^{n+1}$

Most of these terms nearly match the corresponding terms of $S$.  For example, $S$ has the term $3q^{3}$ while $qS$ has the term $2q^{3}$.  We need one more $q^{3}$ in the new series.  We also need a $q$, one more $q^{2}$, one more $q^{4}$, and so on.  In short, we need to add $Q$ to the new series.  This gets us a perfect match for all of the terms of $S$, with an additional term: $qS+Q=q+2q^{2}+3q^{3}+4q^{4}+...+nq^{n}+nq^{n+1}$

Subtracting $S$ from this, almost everything cancels out: $qS+Q-S=nq^{n+1}$, $S=\dfrac {nq^{n+1}-Q} {q-1}$.

Applying Formula 4.4 and a little more algebra, we find the shortcut formula for $S$: $S=\dfrac {q-(n+1)q^{n+1}+nq^{n+2}} {(q-1)^{2} }$                                                                                 (Formula 5.1)

We could also find Formula 5.1 by exploiting some basic calculus.  You might have noticed that $S$ is very similar to the derivative of the geometric series $Q$: $Q'=1+2q+3q^{2}+...+{n}q^{n-1}$

Multiplying by $q$ we see that $S=qQ'$.  So all we need is a short formula for $Q'$.  By Formula 4.4, we can find $Q'$ by taking the derivative of $\dfrac {q-q^{n+1}}{1-q}$, using a rule for the derivative of the quotient.  If you work it out, you will end up with Formula 5.1.

Formula 5.1 is the other trick I used to find Formula 3.1.

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