deletions | additions       diff --git a/The 21 thing.tex b/The 21 thing.tex  index 60e7335..1ac65bc 100644  --- a/The 21 thing.tex  +++ b/The 21 thing.tex 
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\sum_{n=1}^{\infty} \frac{ (21;;1)^n }{ (21;;2) ^n} = \frac{1}{10}\frac{1}{(10;;1)} \\  \sum_{n=1}^{\infty} \frac{ (21;;1)^n }{ (21;;3) ^n} = \frac{1}{10}\frac{1}{(10;;2)} \\  \sum_{n=1}^{\infty} \frac{ (21;;1)^n }{ (21;;4) ^n} = \frac{1}{10}\frac{1}{(10;;3)} \\  \sum_{n=1}^{\infty} \frac{ (21;;1)^n }{ (21;;g) ^n} = \frac{1}{10}\frac{1}{(10;;g-1)} \\  \sum_{n=1}^{\infty} \frac{  (21;;2)^n }{ (21;;3) ^n} = \frac{101}{10000} \frac{1}{1000}\frac{101}{10} \\  \sum_{n=1}^{\infty} \frac{ (21;;2)^n }{ (21;;4) ^n} = \frac{1}{1000}\frac{101}{1010} \\  \sum_{n=1}^{\infty} \frac{ (21;;2)^n }{ (21;;5) ^n} = \frac{1}{1000}\frac{101}{101010} \\  \sum_{n=1}^{\infty} \frac{ (21;;2)^n }{ (21;;g) ^n} = \frac{1}{1000}\frac{101}{(10;;g-2)} \\  \sum_{n=1}^{\infty} \frac{ (21;;3)^n }{ (21;;4) ^n} = \frac{1}{100000}\frac{10101}{10} \\  \sum_{n=1}^{\infty} \frac{ (21;;3)^n }{ (21;;5) ^n} = \frac{1}{100000}\frac{10101}{1010} \\  \sum_{n=1}^{\infty} \frac{ (21;;3)^n }{ (21;;g) ^n} = \frac{1}{100000}\frac{10101}{(10;;g-3)} \\  \end{equation}  By which one can conclude \begin{equation}  \sum_{n=1}^{\infty} \frac{ (21;;f)^n }{ (21;;g) ^n} = \frac{1}{10^{2f-1}}\frac{1|(01;;f-1)}{(10;;g-f)}  \end{equation}