No no Please not Double Factorials

The "last" term of this sum is $\dfrac{4017!!}{4018!!} = \dfrac{4017*4015*4013* ....*3*1}{4018*4016*4014*....*4*2}.$

When we reduce this fraction, for the denominator we will be left with all powers of $2$ extracted from the terms in the denominator, as all the odd powers from these terms will be cancelled out by "matching" terms in the numerator. So the denominator will be $2$ raised to the power

$\displaystyle\sum_{k=1}^{11} \lfloor\dfrac{4018}{2^{k}} \rfloor = 4010.$

All other terms in the original sum can be treated in the same way, with denominators of the form $2^{m}$ for $m \lt 4010.$ Thus when we add all the terms together the least common denominator will be $2^{4010},$ and so

$\dfrac{ab}{10} = \dfrac{4010*1}{10} = \boxed{401}.$