If the middle terms of  ${(1+x)}^{2n+1}$ are $\frac{\mathrm{1.3.5}\cdots (2n+1)k}{(n+1)!}\text{\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{1.3.5}\cdots (2n+1)kx}{(n+1)!}$  then $k=$

${(1+x)}^{2n+1}(2n+1)=\text{odd}$

Middle terms $\frac{T_{2n+1}+1}{2},\frac{T_{2n+1}+3}{2}$

$\begin{array}{l}{T}_{n+1},T_{n+2}\\ T_{n+1}={}^{2n+1}{\text{C}}_n{x}^n\\ T_{n+2}={}^{2n+1}{\text{C}}_{n+1}{x}^{n+1}\\ {\text{given\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}}^{2n+1}{\text{C}}_n{x}^n=\frac{\mathrm{1.3.5}\cdots (2n+1)k}{(n+1)!}\\ \frac{(2n+1)!x^n}{(n+1)!n!}=\frac{\mathrm{1.3.5}\cdots (2n+1)k}{(n+1)!}\\ \frac{(2n+1)!\left(2n\right)!x^n}{n!}=\mathrm{1.3.5}\cdots (2n+1)k\\ \frac{(2n+1)\left(2^n\right)n![\mathrm{1.3.5}\cdots (2n-1)]x^n}{n!}=\mathrm{1.3.5}\cdots (2n+1)k\\ 2^n.x^n=k\\ k={\left(2x\right)}^n\end{array}$