$\sum _{k=1}^{2015}\frac{\left(k+2016\right)!}{\left(k-1\right)!}=\frac{\left(A\right)!}{\left(B\right)!\left(C\right)}$

Find the value of $A+B+C$

The answer is 8064.

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The sum is $2017! \sum_{k=1}^{2015} \binom{k+2016}{2017},$ which equals $2017! \binom{4032}{2018} = \frac{4032!}{(2014!)2018},$ so the answer is $4032+2014+2018 = \fbox{8064}.$