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∑n=12015n(n+2)(n+3)(n+4)=AB\large{\displaystyle \sum^{2015}_{n=1}\frac{n}{(n+2)(n+3)(n+4)}=\frac{A}{B}}n=1∑2015(n+2)(n+3)(n+4)n=BA
Find A+BA+BA+B where A,BA,BA,B are positive coprime integers.
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