$\text{ Let }{\left(1+x+2x^2\right)}^{20}=a_0+a_{1}x+a_2{x}^2+\dots +a_{40}{x}^{40}\text{ . Then }{a}_1+a_3+a_5+\dots .+a_{37}\text{ is equal to }$

${\left(1+x+2x^2\right)}^{20}=a_0+a_{1}x+\mathrm{.....}{a}_{40}{x}^{40}$

$x=1\Rightarrow 4^{20}=a_0+a_1+\dots \dots ..+a_{40}$

$x=-1\Rightarrow 2^{20}=a_0-a_1+a_2\dots \dots .+a_{40}$

$\frac{4^{20}-2^{20}}{20}=a_1+a_3+\dots \dots a_{37}+a_{39}$

$\text{ But }{a}_{39}=\text{ coeff of }{x}^{39}in {\left(1+x+2x^2\right)}^{20}$

$=\frac{20!}{p!q!r!}{1}^p.x^q{.2}^r.x^{2r}$

$q+2r=39,p+q+r=20$

$q=1,r=19,p=0$

$a_{39}=\frac{20!}{0!1!19!}{.2}^{19}={20.2}^{19}$

$\text{ Required }=\frac{2^{40}-2^{20}}{2}-20.2^{19}$

$=\frac{2^{40}-2^{20}-{20.2}^{20}}{2}$

$=2^{39}-2^{19}-20.2^{19}=2^{19}\left(2^{20}-21\right)$