Super Modified Combination Sum

The chairman identity tells us that $r \binom{n}{r} = n \binom {n-1}{r-1}$ . Using that we get $\sum_{k=1}^{504} (4k-1) \binom{2015}{4k-1} =\sum_{k=1}^{504} (2015) \binom{2014}{4k-2}$

Now, we can factor out the $2015$ and multiply it back on later. We now need to evaluate $\sum_{k=1}^{504} \binom{2014}{4k-2}$ . From the binomial expansion of $(1+(-1))^n$ , we see that $\sum_{k=1}^{\frac{n}{2}} \binom{n}{2k-1}=\sum_{k=1}^{\frac{n}{2}} \binom{n}{2k}=\frac {1}{2} \sum_{k=0}^{n} \binom{n}{k}$

Now looking at the sum we have, we can see that $\binom{2014}{2}=\binom{2014}{2012}, \binom{2014}{6}=\binom{2014}{2008}, \ldots, \binom{2014}{2014}=\binom{2014}{0}$

Thus, we have $\sum_{k=1}^{504} \binom{n}{4k-2}= \sum_{k=1}^{504} \binom{n}{4k-4}=\frac{1}{2} \sum_{k=0}^{1007} \binom{2004}{2k}=\frac{1}{4} \sum_{k=0}^{2014} \binom{2014}{k}$

Since we have $\sum_{k=0}^{2014} \binom{2014}{k}=2^{2014}$

It follows that $\sum_{k=1}^{504} \binom{n}{4k-2}=\frac {1}{4} 2^{2014}=2^{2012}$

Now we multiply the $2015$ back on to get the desired sum of ${2015(2^{2012})}$

$\therefore$ Our answer is $2015+2+2012=\boxed{4029}$

$\begin{aligned} (1+x)^{2015} & = {2015 \choose 0} + {2015 \choose 1}x + ... + {2015 \choose 2015}x^{2015} \\ 2015(1+x)^{2014} & = {2015 \choose 1} + 2{2015 \choose 2}x + 3{2015 \choose 3}x^2 + ... + 2015{2015 \choose 2015}x^{2014} \\ f(x) = 2015x^2(1+x)^{2014} & = {2015 \choose 1}x^2 + 2{2015 \choose 2}x^3 + ... + 2015{2015 \choose 2015}x^{2016} \end{aligned}$ So the required expression is $\frac{f(i) + f(-1) + f(-i) + f(1)}{4} = 2015 \cdot 2^{2012} .$

$\sum _{k=1}^{504} (4 k-1) \binom{2015}{4 k-1}$ is $2^{2012}\times 5\times 13\times 31$ . Since $2$ is the only prime with an exponent greater than 1, $p=2105, q=2, r=2012$ .