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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.