Are you crazy? It's 2016

There is also a nice combinatorial way of interpreting the identity.

Consider a class of 2016 students. Our aim is to select any group of students and then assign 3 different jobs to the group such that one job is allocated to one person only. Note that it is possible that multiple jobs are assigned to the same person. The sum $A$ in the above question represents the obvious way of finding the answer. Another way to solve the counting problem is to first allocate the 3 jobs to the students and then forming a group.

Making 3 cases based on the number of students to which a job is assigned (either 1,2 or 3 students) the answer can be easily obtained as

$\large A= (2016)(2015)(2014)2^{2013}+(2016)(2015)(3)(2^{2014})+(2016)2^{2015}$

We use generating functions to solve this.

Consider $F(x) = (1+x)^{2016} = \binom{2016}{0} + \binom{2016}{1} x^1 + \cdots + \binom{2016}{2016} x^{2016}$

It is not hard to see that $G(x) = \frac{d}{dx}[x \cdot \frac{d}{dx} [x \cdot \frac{d}{dx} F(x)]] = \binom{2016}{1} \cdot 1^3 \cdot x^0 + \binom{2016}{2} \cdot 2^3 \cdot x^1 + \cdots + \binom{2016}{2016} \cdot 2016^3 \cdot x^{2015}$

After a bit of differentiation and algebra, we find $G(x) = 2016(1+x)^{2015} + 2016 \cdot 2015x(1+x)^{2014} + 2016 \cdot 2015 [2x(1+x)^{2014} + 2014x^2(1+x)^{2013}]$

We want the digit sum of $\frac{A}{2016 \cdot 2^{2016}} = \frac{G(1)}{2016 \cdot 2^{2016}} = \frac{2016 \cdot 2^{2014} [2 \cdot 2016 + 2015 + 2015 \cdot 1007]}{2016 \cdot 2^{2016}} = \frac{2016 \cdot 2^{2016} \cdot 508788}{2016 \cdot 2^{2016}} = 508788$ which is just $\boxed{36}$

Where have u copied this problem from? Plz learn to give credit man..... ;)