Question

\(N=\dfrac{1.3.5...2017.2019}{1011.1012.1013...2019.2020}\)

Answer 1

\(N=\dfrac{\left(2\cdot4\cdot6\cdot...\cdot2020\right)\cdot\left(1\cdot3\cdot5\cdot...\cdot2017\cdot2019\right)}{\left(2\cdot4\cdot6\cdot...\cdot2020\right)\cdot\left(1011\cdot1012\cdot...\cdot2019\cdot2020\right)}\)

\(=\dfrac{1\cdot2\cdot3\cdot...\cdot2020}{\left[\left(1.2\right)\cdot\left(2\cdot2\right)\cdot...\left(2\cdot1010\right)\right]\cdot\left[1011\cdot1012\cdot...\cdot2019\cdot2020\right]}\)

=1/2^1010