The coefficient of middle term in the expansion of $(1+{x})^{40}$ is

${\textbf{Step - 1: Determine the Middle term}}$

${\text{Middle term = }}\dfrac{{\text{N}}}{2} + 1 = \dfrac{{2{\text{n}}}}{2} + 1$

$= \left( {{\text{n + 1}}} \right){\text{th term}}$

${\text{The middle term of }}{\left( {{\text{1 + x}}} \right)^{40}}{\text{ is }}\left( {\dfrac{{40}}{2} + 1} \right){\text{ i}}{\text{.e 2}}{{\text{1}}^{th}}{\text{ term in that expansion}}$

${{\text{T}}_{21}} = {{\text{ }}^{40}}{{\text{C}}_{20}}{{\text{x}}^{40 - 20}} = {{\text{ }}^{40}}{{\text{C}}_{20}}{{\text{x}}^{20}}$

${\text{Therefore the coefficient of the middle term is}}{{\text{ }}^{40}}{{\text{C}}_{20}}$

$\therefore {{\text{ }}^{40}}{{\text{C}}_{20}} = \dfrac{{40!}}{{20!\left( {40 - 20} \right)!}} = \dfrac{{40!}}{{20!20!}}$

${\textbf{Step - 2 : Simplify the term}}{\text{.}}$

${\text{Dividing by 20!;and separating the multiples of two in the powers of two we get,}}$

$\therefore {\text{Middle term = }}\dfrac{{\left( {1.3.5.7.....39} \right){2^{20}}}}{{20!}}$

${\textbf{Hence, Correct answer will be }}\dfrac{{\left( {1.3.5.7.....39} \right){2^{20}}}}{{20!}}$