Question:
Let $\left(1+x+2 x^{2}\right)^{20}=a_{0}+a_{1} x+a_{2} x^{2}+\ldots+a_{40} x^{40}$ then $\mathrm{a}_{1}+\mathrm{a}_{3}+\mathrm{a}_{5}+\ldots+\mathrm{a}_{37}$ is equal to
Correct Option: , 2
Solution:
$\left(1+x+2 x^{2}\right)^{20}=a_{0}+a_{1} x+\ldots+a_{40} x^{40}$ put $\mathrm{x}=1,-1$
$\Rightarrow a_{0}+a_{1}+a_{2}+\ldots+a_{40}=2^{20}$
$\Rightarrow a_{1}+a_{3}+\ldots+a_{39}=\frac{4^{20}-2^{20}}{2}$
$\Rightarrow a_{1}+a_{3}+\ldots+a_{37}=2^{39}-2^{19}-a_{39}$
here $a_{39}=\frac{20 !(2)^{19} \times 1}{19 !}=20 \times 2^{19}$
$\Rightarrow a_{1}+a_{3}+\ldots+a_{37}=2^{19}\left(2^{20}-1-20\right)$
$=2^{19}\left(2^{20}-21\right)$