Welcome 2017 !

The sum $(\mathfrak{S})$ is: $\sum_{k=4}^{2016}\left[(-1)^k\dfrac{ (k+1)^3}{k((k-1)!+k!+(k+1)!)}\right]$ The denominator is: $\small{ k((k-1)!+k!+(k+1)!)=k(k-1)!(1+k+k(k+1))=k!(k^2+2k+1)=k!(k+1)^2}$

$\therefore \mathfrak{S}= \sum_{k=4}^{2016}\left[(-1)^k\dfrac{k+1}{k!}\right]=\sum_{k=4}^{2016}(-1)^k\left[\dfrac{1}{(k-1)!}+\dfrac{1}{k!}\right]$

Thus expanding, we get:

$\mathfrak S=\dfrac{1}{3!}+\cancel{\dfrac{1}{4!}}\color{#3D99F6}{-\cancel{\dfrac{1}{4!}}-\cancel{\dfrac{1}{5!}}}\color{#333333}{+\cancel{\dfrac{1}{5!}}+\cancel{\dfrac{1}{6!}}}+\cdots\color{#3D99F6}{+\cancel{\dfrac{1}{2015!}}+\dfrac{1}{2016!}}$

$\large =\dfrac{1}{3!}+\dfrac{1}{2016!}$

$\therefore a=3,b=2016,c=0,d=-1\implies a+b+c+d=\boxed{\color{cyan}{2018}}$