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∑n=12015n2+n+1(n2+n)(n+1)!\displaystyle{\sum _{ n=1 }^{ 2015 }{ \frac { { n }^{ 2 }+n+1 }{ \left( { n }^{ 2 }+n \right) \left( n+1 \right) ! } }}n=1∑2015(n2+n)(n+1)!n2+n+1 can be represented in the form ab\dfrac { a }{ b } ba, where aaa and bbb are coprime positive integers. Find the value of a−ba-ba−b.
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