Chaitanya's Random Pages

September 2, 2011

Addendo to sum geometric series

Filed under: mathematics — ckrao @ 12:00 pm

In an earlier post on manipulating fractions, I mentioned the useful addendo property: if a/b = c/d, then

\displaystyle \frac{a}{b} = \frac{c}{d} = \frac{a+c}{b+d}.

This property is in fact used in Euclid’s Elements (Book 9, Proposition 35) written some 2300 years ago to sum a geometric series!

The usual way one is taught to find the sum

\displaystyle S = a + ar + ar^2 + ar^3 + \ldots + ar^{n-1}

is to multiply both sides by r to give

\displaystyle Sr = ar + ar^2 + ar^3 + ar^4 + \ldots + ar^n

and then subtract the first equation from the second:

\displaystyle S(r-1) = ar^n - a.

Hence \displaystyle S = \frac{a(r^n - 1)}{r-1}.

Here is an alternative approach based on Euclid’s work. We begin with the ratios

\displaystyle \frac{ar}{a} = \frac{ar^2}{ar} = \frac{ar^3}{ar^2} = \ldots = \frac{ar^n}{ar^{n-1}}.

Subtracting 1 from each of these ratios (equal to r) gives

\displaystyle \frac{ar - a}{a} = \frac{ar^2-ar}{ar} = \frac{ar^3-ar^2}{ar^2} = \ldots = \frac{ar^n-ar^{n-1}}{ar^{n-1}}.

By addendo each of these ratios (equal to r-1) is also equal to

\displaystyle \frac{(ar - a) + (ar^2 - ar) + (ar^3 - ar^2) + \ldots + (ar^n - ar^{n-1})}{a + ar + ar^2 + \ldots + ar^{n-1}}.

The numerator here simplifies greatly and we are left with

\displaystyle r-1 = \frac{ar^n - a}{a + ar + ar^2 + \ldots + ar^{n-1}},

from which

\displaystyle a + ar + ar^2 + \ldots + ar^{n-1} = \frac{a(r^n - 1)}{r-1}

as desired.

Geometric sequences have a special self-similarity property – multiplying each term by the common ratio r produces a new sequence that only differs from the original sequence at its ends. Keeping this idea in mind makes them easy to sum.

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