Just too free these days 2

We write the general case for the sum upto $a$ terms:

$\displaystyle \Rightarrow \displaystyle \sum_{n=1}^{a} \frac{(n^2 + n + 1)(n^2 - n +1)}{n(n+1)(n^2 - n + 1)}$

$\displaystyle \Rightarrow \displaystyle \sum_{n=1}^{a} 1 + \frac{1}{n(n + 1)}$

$\displaystyle \Rightarrow a + \displaystyle \sum_{n=1}^a \frac{1}{n} - \frac{1}{n+1}$

$\displaystyle \Rightarrow a + 1 - \frac{1}{a + 1}$

$\displaystyle \Rightarrow \frac{(a + 1)^2 -1}{(a+1)}$

We know that for a natural number $k$ , $k^2 - 1$ and $k$ are co-prime. Hence, we conclude:

$\displaystyle \frac{(a+1)^2 - 1}{a+1} = \frac{A}{B}$

$\displaystyle \Rightarrow A + B = (a + 1)^2 + (a + 1) - 1$

Substituting $a = 2016$ , we get:

$\displaystyle A + B = \boxed{4070305}$

The summation can be simplified as $\sum _{ n=1 }^{ 2016 }{ \frac { { n }^{ 2 }+n+1 }{ { n }^{ 2 }+n } }$