Find the remainder when $\sum\limits_{r = 0}^{2014} {\sum\limits_{k = 0}^r {{{\left( { - 1} \right)}^k}.\left( {k + 1} \right)} } \left( {k + 2} \right).{}^{2019}{C_{r - k}}$ is divisible by 64.
Note: $^n{C_r} = \frac{{n!}}{{r!\left( {n - r} \right)!}}$
Not able to solve this problem at the initial sum is upto 2014 where as in the end the series used its summation is up-to 2019,
${\sum\limits_{k = 0}^r {{{\left( { - 1} \right)}^k}.\left( {k + 1} \right)} } \left( {k + 2} \right).{}^{2019}{C_{r - k}}$ can be evaluated using the identity
${\sum\limits_{k = 0}^r {{{\left( { - 1} \right)}^k}.\left( {k + 1} \right)} } \left( {k + 2} \right).{}^{n}{C_{r - k}}$ = $ {}^{n}{C_{r}} \frac{(n-r)2r^2+(6-4n)r+(2n^2-6n+4)}{n(n-1)(n-2)}$
The sum finally ends up with $\sum\limits_{r = 0}^{2014} {}^{n}{C_{r}} \frac{(n-r)2r^2+(6-4n)r+(2n^2-6n+4)}{n(n-1)(n-2)}$