Question

If ${\left(1+x-3x^2\right)}^{10}=1+a_{1}x+a_2{x}^2+...+a_{20}{x}^{20}$, then $a_2+a_4+a_6+...+a_{20}$ is equal to

• A. $\frac{3^{10}+1}{2}$
• B. $\frac{3^9+1}{2}$
• C. $\frac{3^{10}-1}{2}$
• D. $\frac{3^9-1}{2}$

Answer 1

Answer: (C)

${\left(1+x-3x^2\right)}^{10}=1+a_{1}x+a_2{x}^2+a_3{x}^3+...+a_{20}{x}^{20}$
Putting $x=1$, we get
${\left(1+1-3\right)}^{10}=1+a_1+a_2+a_3+...+a_{20}$
$\therefore 1=1+a_1+a_2+a_3+...+a_{20}\dots \left(i\right)$
Putting $x=-1$, we get
${\left(1-1-3\right)}^{10}=1-a_1+a_2-a_3+...+a_{20}$
$\Rightarrow {\left(3\right)}^{10}=1-a_1+a_2-a_3+...+a_{20}\dots \left(ii\right)$
Adding $\left(i\right)$ and $\left(ii\right)$, $3^{10}+1=2\left(1+a_2+a_4+.....+a_{20}\right)$
$\therefore a_2+a_4+a_6+...+a_{20}=\frac{3^{10}+1-2}{2}=\frac{3^{10}-1}{2}$