If ${\left(1+x+2x^2\right)}^{20}=a_0+a_{1}x+a_2{x}^2+\dots +a_{40}{x}^{40}$ then $a_1+a_3+a_5+\dots +a_{37}$ equals

Putting $x=1$ and -1 and subtracting, we get

$\begin{array}{l}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{4}^{20}-2^{20}=2\left[a_1+a_3+a_5+\dots +a_{39}\right]\\ \text{ or }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{2}^{19}\left(2^{20}-1\right)=\left[a_1+a_3+\dots +a_{37}\right]+a_{39}\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}E=2^{19}\left(2^{20}-1\right)-a_{39 } ....\left(1\right)\end{array}$

Now we have to find $a_{39}$ which is coefficient of $x^{39}$ in 

$\begin{array}{l}{\left(1+x+2x^2\right)}^{20}\text{ i.e., in }\{1+x(1+2x){\}}^{20}\\ T_{r+1}{=}^{20}{C}_r{x}^T(1+2x{)}^r\end{array}$

Put $r=20$ so that 

$T_{21}{=}^{20}{C}_{20}{x}^{20}(1+2x{)}^{20}=x^{20}(1+2x{)}^{20}$     ….(2)

Now we will find the coefficient of $x^{19}\text{ in }(1+2x{)}^{20}$

   which we know is

${ }^{20}{C}_{19}(2x{)}^{19}={20.2}^{19}{x}^{19}$

$\therefore$ The coefficient of $x^{39}\text{ in }\left(1\right)\text{ is }{20.2}^{19}$

    Putting in (1), we get

$E=2^{19}\left(2^{20}-1\right)-20\cdot 2^{19}=2^{19}\left(2^{20}-21\right)$