jmc

Note that $2014\equiv -3\mathrm{mod}2017$. We have for $k\ge 1$$$\left(\genfrac{}{}{0ex}{}{2014}{k}\right)\equiv \frac{(-3)(-4)(-5)....(-2-k)}{k!}\mathrm{mod}2017$$$$\equiv (-1{)}^k(\genfrac{}{}{0ex}{}{k+2}{k})\mathrm{mod}2017$$$$\equiv (-1{)}^k(\genfrac{}{}{0ex}{}{k+2}{2})\mathrm{mod}2017$$Therefore$$\sum _{k=0}^{62}\left(\genfrac{}{}{0ex}{}{2014}{k}\right)\equiv \sum _{k=0}^{62}(-1{)}^k(\genfrac{}{}{0ex}{}{k+2}{2})\mathrm{mod}2017$$This is simply an alternating series of triangular numbers that goes like this: $1-3+6-10+15-\mathrm{21....}$ After finding the first few sums of the series, it becomes apparent that$$\sum _{k=1}^n(-1{)}^k(\genfrac{}{}{0ex}{}{k+2}{2})\equiv -\left(\frac{n+1}{2}\right)\left(\frac{n+1}{2}+1\right)\mathrm{mod}2017\text{ if n is odd}$$and$$\sum _{k=1}^n(-1{)}^k(\genfrac{}{}{0ex}{}{k+2}{2})\equiv {\left(\frac{n}{2}+1\right)}^2\mathrm{mod}2017\text{ if n is even}$$Obviously, $62$ falls in the second category, so our desired value is$${\left(\frac{62}{2}+1\right)}^2=32^2=\boxed{1024}$$