The remainder of the year, corrected.

Firstly we can use the identity that $\sum_{i=0}^n (i^2) = \dbinom{2n}{n}$ . However the sum in the question begins at $i=1$ , so we have to subtract one. So the question now is: Find $(4028!) \div (2014!)(2014!) -1 \pmod{53}$ . Now, $2014 \div 53 = 38$ so $53^{76}$ is the highest power of 53 which divides the denominator. $4028 \div 53 = 76$ , however one of these multiples of 53 is $53^{2}$ so in fact $53^{77}$ is the highest power of 53 which divides the numerator. So $53 \mid (4028!)\div(2014!)(2014!)$ , and therefore $(4028!) \div (n!)(n!) - 1 \equiv 0-1 \equiv 52 \pmod{53}$