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Geometric Series and Positional Notation

By: Gijs Bellaard

A few nights ago I thought of the following cute little proof. Lets start with something easy: \[ 999 = 1000 - 1\] Dividing both sides by \(9 = 10 - 1\): \[ 111 = \frac{1000 - 1}{10-1}\] So, in general: \[ 10^n + 10^{n-1} + \dots + 1 = \frac{10^{n+1} - 1}{10-1} \] But, interestingly, every step we did works in any base \(q\), and therefor even more generally: \[ q^n + q^{n-1} + \dots + 1 = \frac{q^{n+1} - 1}{q-1} \] Which is the well-know formula for the sum of the first \(n\) terms of a geometric series.