Binomial Coefficients modulo $2017$

Since 2014 is so close to 2017, we may think to write out the formula for the binomial coefficients directly $\pmod{2017}$ : $\dbinom{2014}{k} \equiv \dfrac{(-3)\times(-4)\times...\times(-(k+2))}{k!} \pmod{2017}$ So in fact almost all of the terms will cancel out, leaving us with: $\dbinom{2014}{k} \equiv \dfrac{(k+1)(k+2)}{2} \pmod{2017}$ if k is even and $\dbinom{2014}{k} \equiv - \dfrac{(k+1)(k+2)}{2} \pmod{2017}$ if k is odd. Therefore $\displaystyle\sum_{n=0}^{62} \dbinom{2014}{n} \equiv 1-3+6-10+...-\dfrac{62 \times 63}{2}+\dfrac{63 \times 64}{2} \pmod{2017}$ . Considering the consecutive pairs of terms we can evaluate this by noting that it is just an arithmetic series with 31 terms, first term -2 and common difference -2. We must also remember to add on the final term $\dfrac{63 \times 64}{2}$ which is not in a pair. As a side note, again to evaluate all the terms except the final one, there is a beautiful proof without words here: http://www.maa.org/sites/default/files/Plaza2007-236990.pdf

However we evaluate it, we get that $\displaystyle\sum_{n=0}^{62} \dbinom{2014}{n} \equiv 1024 \pmod{2017}$ so $a = 32$