Let (1 + x + 2 x ^ 2 ) ^ 20 = a _ 0 + a _ 1 x + a _ 2 x ^ 2 + + a _...

Given ${\left(1 + x + 2{x}^{2}\right)}^{20} = {a}_{0} + {a}_{1}x + \ldots + {a}_{40}{x}^{40}$

Put $x =$$1$ we get,

${a}_{0} + {a}_{1} + {a}_{2} + \ldots + {a}_{40} = {4}^{20}$

Put $x = - 1$ we get,

${a}_{0} - {a}_{1} + {a}_{2} + \ldots . + {a}_{40} = {2}^{20}$

$⇒ {a}_{1} + {a}_{3} + \ldots . + {a}_{39} = \dfrac{{4}^{20} - {2}^{20}}{2}$

$⇒ {a}_{1} + {a}_{3} + \ldots + {a}_{37} = {2}^{39} - {2}^{19} - {a}_{39}$

Here ${a}_{39} = \dfrac{20 ! ( 2{)}^{19} \times 1}{19 !} = 20 \times {2}^{19}$

$⇒ {a}_{1} + {a}_{3} + \ldots + {a}_{37} = {2}^{19}\left({2}^{20} - 1 - 20\right)$

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