Question

If ${(1+x+2x^2)}^{20}=a_0+a_{1}x+a_2{x}^2+....+a_{39}{x}^{39}+a_{40}{x}^{40}$ then find the value of $a_0+a_1+a_2+......+a_{38}$

• A. $2^{19}(2^{20}-1)$
• B. $2^{19}(2^{20}+1)$
• C. $2^{20}(2^{19}-1)$
• D. $2^{20}(2^{19}+1)$

Answer 1

The correct option is A $2^{19}(2^{20}-1)$

${(1+x+{2x}^2)}^{20}=a_0+a_{1}x+a_2{x}^2+........+a_{40}{x}^{40}$ ....(1)
Put $x=1$ in (1)
$4^{20}=a_0+a_1+a_2+........+a_{40}$
$\Rightarrow 2^{40}=a_0+a_1+a_2+........+a_{40}$ ....(2)
Put $x=-1$ in (1)
$\Rightarrow 2^{20}=a_0-a_1+a_2-a_3........-a_{39}+a_{40}$ ....(3)
Adding (2) and (3), we get
$2^{40}+2^{20}=2(a_0+a_2+a_4+......+a_{38}+a_{40})$
$2^{20}(2^{20}+1)=2(a_0+a_2+a_4+......+a_{38}+a_{40})$
$\Rightarrow 2^{19}(2^{20}+1)=a_0+a_2+a_4+......+a_{38}+a_{40}$
$a_0+a_2+a_4+......+a_{38}=2^{19}(2^{20}+1)-a_{40}$ .....(4)
Since, $a_{40}$ is the coefficient of $x^{40}$ in the expansion of $(1+x+2x^2{)}^{20}$
$\Rightarrow a_{40}=2^{20}$
Put this value in (4), we get
$a_0+a_2+a_4+......+a_{38}=2^{19}(2^{20}+1)-2^{20}$
$\Rightarrow a_0+a_2+a_4+......+a_{38}=2^{19}(2^{20}-1)$