Why do I love 2015 so much #2

$\begin{aligned} \frac{1}{n+2} + \frac{1}{n+3} - \frac{2}{n+4} & = \frac{3n+8}{(n+2)(n+3)(n+4)} \\ \Rightarrow \frac{3n}{(n+2)(n+3)(n+4)} & = \frac{1}{n+2} + \frac{1}{n+3} - \frac{2}{n+4} - \frac{8}{(n+2)(n+3)(n+4)} \\ & = \frac{1}{n+2} + \frac{1}{n+3} - \frac{2}{n+4} - 4 \left(\frac{1}{n+2} - \frac{2}{n+3} + \frac{1}{n+4} \right) \\ & = - \frac{3}{n+2} + \frac{9}{n+3} - \frac{6}{n+4} \\ \frac{n}{(n+2)(n+3)(n+4)} & = - \frac{1}{n+2} + \frac{3}{n+3} - \frac{2}{n+4} \\ \Rightarrow \sum_{n=1}^{2015} \frac{n}{(n+2)(n+3)(n+4)} & = \sum_{n=1}^{2015} \left( - \frac{1}{n+2} + \frac{3}{n+3} - \frac{2}{n+4} \right) \\ & = - \sum_{n=3}^{2017} \frac{1}{n} + \sum_{n=4}^{2018} \frac{1}{n} + 2\sum_{n=4}^{2018} \frac{1}{n} - 2\sum_{n=5}^{2019} \frac{1}{n} \\ & = - \frac{1}{3} + \frac{1}{2018} + \frac{2}{4} - \frac{2}{2019} \\ & = \frac{112840}{679057} \\ \\ \Rightarrow A+B & = 112840 + 679057 \\ & = \boxed{791897} \end{aligned}$

This can be written as

$\frac { 1 }{ 2 } \sum _{ n=1 }^{ 2015 }{ \frac { 2n+4-4 }{ (n+2)(n+3)(n+4) } }$

$\frac { 1 }{ 2 } \left( \sum _{ n=1 }^{ 2015 }{ \frac { 2(n+2) }{ (n+2)(n+3)(n+4) } } -\frac { 1 }{ 2 } \frac { 4+(n+4)-(n+2) }{ (n+2)(n+3)(n+4) } \right)$

$\sum _{ n=1 }^{ 2015 }{ \left( \frac { 1 }{ (n+3) } -\frac { 1 }{ (n+4) } \right) } -\sum _{ n=1 }^{ 2015 }{ \left( \frac { 1 }{ (n+2)(n+3) } -\frac { 1 }{ (n+3)(n+4) } \right) }$

So, have expressed the sum as difference of consecutive terms. All the intermediate terms get cancelled and only first and last terms are left.

So the answer is $\frac { 1 }{ 6 } -\frac { 2017 }{ 2018\times 2019 }$ . On simplifying we get $A+B=791897$