Lawrence's sum

Algebra Level 4

If the sum

n=12111n+2n!+(n+1)!+(n+2)! \sum_{n=1}^{2111} \frac{n+2}{n! + (n+1)! + (n+2)!}

is written as 121a! \frac{1}{2} - \frac{1}{a!}, what are the last three digits of aa?

This problem is posed by Lawrence L.

Details and assumptions

The last three digits of 10231023 is 023023. You may enter your answer as 023023 or 2323.


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