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If the sum
∑n=12111n+2n!+(n+1)!+(n+2)! \sum_{n=1}^{2111} \frac{n+2}{n! + (n+1)! + (n+2)!} n=1∑2111n!+(n+1)!+(n+2)!n+2
is written as 12−1a! \frac{1}{2} - \frac{1}{a!}21−a!1, what are the last three digits of aaa?
This problem is posed by Lawrence L.
Details and assumptions
The last three digits of 102310231023 is 023023023. You may enter your answer as 023023023 or 232323.
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