Let left(1+x+2 x^{2}right)^{20}=a_{0}+a_{1} x | vedclass

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Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

$\left(1+x+2 x^{2}ight)^{20}=a_{0}+a_{1} x+\ldots .+a_{40} x^{40}$ put $x=$ $1,-1$

$\Rightarrow a_{0}+a_{1}+a_{2}+\ldots .+a_{40}=2^{20}$

$a_{0}-a_{1}+a_{2}+\ldots+a_{40}=2^{20}$

$\Rightarrow a_{1}+a_{3}+\ldots+a_{39}=\frac{4^{20}-2^{20}}{2}$

$\Rightarrow a_{1}+a_{3}+\ldots+a_{37}=2^{39}-2^{19}-a_{39}$

here $a_{39}=\frac{20 !(2)^{19} 	imes 1}{19 !}=20 	imes 2^{19}$

$\Rightarrow a_{1}+a_{3}+\ldots+a_{37}=2^{19}\left(2^{20}-1-20ight)$

$=2^{19}\left(2^{20}-21ight)$

Let $\left(1+x+2 x^{2}\right)^{20}=a_{0}+a_{1} x+a_{2} x^{2}+\ldots+a_{40} x^{40}$ then $a _{1}+ a _{3}+ a _{5}+\ldots+ a _{37}$ is equal to

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