Powers of 2 and the binomial expansion

$(1+e^x)^n=1 +\binom{n}{1} e^x +\binom{n}{2} e^{2x} +\binom{n}{3} e^{3x} +\cdots + \binom{n}{n} e^{nx}$

Differentiate w.r.t $x$

$n (1+e^x)^{n-1} e^x=\binom{n}{1} e^x +2 \binom{n}{2} e^{2x} +3 \binom{n}{3} e^{3x} +\cdots +n \binom{n}{n} e^{nx}$

Again differentiate w.r.t $x$

$n (1+e^x)^{n-1} e^x +n(n-1) (1+e^x)^{n-2} e^{2x}=\binom{n}{1} e^x +2^2 \binom{n}{2} e^{2x} +3^2 \binom{n}{3} e^{3x} +\cdots +n^2 \binom{n}{n} e^{nx}$

As the above is an identity now put $x=0$ then it becomes

$n.2^{n-1} +n(n-1).2^{n-2}=\binom{n}{1} +2^2 \binom{n}{2} +3^2 \binom{n}{3}+\cdots +n^2 \binom{n}{n}$

So our series becomes

$1^2 \binom{n}{1} +2^2 \binom{n}{2} +3^2 \binom{n}{3}+\cdots +n^2 \binom{n}{n}=n.2^{n-1} +n(n-1).2^{n-2}=2n.2^{n-2} + n^2 2^{n-2} - n 2^{n-2} \\ =n.2^{n-2}+n^2.2^{n-2}=n(n+1).2^{n-2}$

Now put $n=2017$ then

$1^2 \binom{2017}{1} +2^2 \binom{2017}{2} +3^2 \binom{2017}{3}+\cdots +2017^2 \binom{2017}{2017}=2017 \times 2018 \times 2^{2017-2}=2017 \times 1009 \times 2^{2016}$

Imagine giving a test to 2017 people, where anyone can pass or fail, and the highest passing scorer and fastest passing test-taker are both given distinction (can be both the same guy). The number of ways to do this if only $i$ people passed the test is $i^2 {2017 \choose i}$ .

Now, the number of ways to do this if any number of people can pass the test is obviously $\large \displaystyle \sum_{i=1}^{2017} i^2 {2017 \choose i}$ . which is the required sum above.

Alternatively, we can just have two cases - one where the highest scorer and fastest taker are different, and one where they are just the same guy. If they were the same guy, there would be $2017$ ways to choose this guy, and since everyone else can pass or fail, everyone else can do so with $2^{2016}$ ways. For the case where they are different, there are just $2017*2016$ ways to choose the two awardees, and $2^{2015}$ ways to make the other test takers pass or fail. So, the number of ways to hold the test is $2017*2^{2016} + 2017*2016*2^{2015} = 2017* 1009 * 2^{2016}$ . It then follows that the sum above equals this number, and so, the largest value of $n$ for which $2^n$ divides this sum is $\boxed{2016}$ .