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31!+2!+3!+42!+3!+4!+53!+4!+5!+…+20011999!+2000!+2001!=ab−cd!\frac {3}{1! + 2! + 3!} + \frac {4}{2! + 3! + 4!} + \frac {5}{3! + 4! + 5!} + \ldots + \frac {2001}{1999!+ 2000! + 2001!} = \frac {a}{b} - \frac {c}{d!}1!+2!+3!3+2!+3!+4!4+3!+4!+5!5+…+1999!+2000!+2001!2001=ba−d!c
The equation above holds true for coprime postive integers aaa and bbb, and ccc and ddd. What is the digit sum of (abcd)(abc)(ab)a(abcd)^{(abc)^{(ab)^a}}(abcd)(abc)(ab)a?
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