Question

If ${(1+x-2x^2)}^{20}=a_0+a_{1}x+a_2{x}^2+\cdots +a_{40}{x}^{40}$ and the value of $a_1+a_3+a_5+\cdots +a_{39}=-2^k$, then $k=$

Answer 1

${(1+x-2x^2)}^{20}=a_0+a_{1}x+a_2{x}^2+\cdots +a_{40}{x}^{40}$

Putting $x=1,$ we get $a_0+a_1+a_2+a_3+\cdots +a_{40}=0\cdots \left(1\right)$

Putting $x=-1,$ we get
$a_0-a_1+a_2-a_3+\cdots -a_{39}+a_{40}=2^{20}\cdots \left(2\right)$

from $\left(1\right)-\left(2\right)$, we get
$2[a_1+a_3+\cdots +a_{39}]=-2^{20}$

$\Rightarrow a_1+a_3+\cdots +a_{39}=-2^{19}\therefore k=19$