Let $\left(1+x+2 x^{2}\right)^{20}=a_{0}+a_{1} x+a_{2} x^{2}+\ldots . .+a_{40} x^{40}$. Then, $a_{1}

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Let $\left(1+x+2 x^{2}ight)^{20}=a_{0}+a_{1} x+a_{2} x^{2}+\ldots . .+a_{40} x^{40}$. Then, $a_{1}+a_{3}+a_{5}+\ldots \ldots . .+a_{37}$ is equal to\

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(c) : Given, $\left(1+x+2 x^{2}ight)^{20}$\ $=a_{0}+a_{1} x+a_{2} x^{2}+\ldots . .+a_{40} x^{40}$\ Putting $x=1$ and $x=-1$, we get $$\begin{equation*}4^{20}=a_{0}+a_{1}+a_{2}+\ldots . .+a_{40} 	ag{i} \end{equation*}$$ and $2^{20}=a_{0}-a_{1}+a_{2}-\ldots \ldots+a_{40}$\ Subtracting (ii) from (i), we get\ $4^{20}-2^{20}=2\left(a_{1}+a_{3}+\ldots+a_{39}ight)$\ $\Rightarrow a_{1}+a_{3}+\ldots+a_{37}+a_{39}=2^{39}-2^{19}$\ Now, $a_{39}=$ Coefficient of $x^{39}$ in $\left(1+x+2 x^{2}ight)^{20}$\ $=\frac{20!}{0!1!19!}(1)^{0}(1)^{1}(2)^{19}=20 \cdot 2^{19}$\ $\Rightarrow \quad a_{1}+a_{3}+\ldots .+a_{37}=2^{39}-2^{19}-20 \cdot 2^{19}$\ $=2^{19}\left(2^{20}-21ight)$\