Still a Messy Close Form?

$\sum_{n=1}^{2015} (-1)^n (\dfrac{n^2}{n!} + \dfrac{n+1}{n!})$

$\sum_{n=1}^{2015} (-1)^n (\dfrac{n}{(n-1)!} + \dfrac{n+1}{n!})$

$-1 - \dfrac{2}{1!} + \dfrac{2}{1!} + \dfrac{3}{2!} - \dfrac{3}{2!} - \dfrac{4}{3!} + ... (-1)^{2015} (\dfrac{2015}{2014!} + \dfrac{2016}{2015!})$

We can see that alternate terms cancel each other

$-1 - \dfrac{2016}{2015!}$

$a + b + c = 4032$

How to do it the other way

notice that if we switch 2015 to 1,3,5,7.... we see

when 2015 is 1 it is $-1-\dfrac{2}{1!}$

when 2015 is 3 it is $-1-\dfrac{4}{3!}$

.....

thus when 2015 is 2015 it is $-1-\dfrac{2016}{2015!})$

so, a=1,b=2016,c=2015,a+b+c= 4032