Solve this following

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 $a_{1}+a_{3}+a_{5}+\ldots+a_{37}$ is equal to

 

  1. $2^{20}\left(2^{20}-21\right)$

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

  3. $2^{19}\left(2^{20}+21\right)$

  4. $2^{20}\left(2^{20}+21\right)$


Correct Option: , 2

Solution:

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

$1,-1$

$\Rightarrow a_{0}+a_{1}+a_{2}+\ldots+a_{40}=2^{20}$

$a_{0}-a_{1}+a_{2}+\ldots+a_{40}=2^{20}$

$\Rightarrow a_{1}+a_{3}+\ldots+a_{39}=\frac{4^{20}-2^{20}}{2}$

$\Rightarrow a_{1}+a_{3}+\ldots+a_{37}=2^{39}-2^{19}-a_{39}$

here $a_{39}=\frac{20 !(2)^{19} \times 1}{19 !}=20 \times 2^{19}$

$\Rightarrow a_{1}+a_{3}+\ldots+a_{37}=2^{19}\left(2^{20}-1-20\right)$

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

 

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