Explanation

(a)
Given , $(1 + x - {2x}^{2})^{6} = 1 + a_{1}x + a_{2}x^{2} + ... + a_{12}x^{12}$ …(i)
On putting $x = 1$ in Eq. (i), we get
$(1 + 1 - 2)^{6} = 1 + a_{1} + a_{2} + \ldots + a_{12}$
$\Longrightarrow {(0)}^{6} = 1 + a_{1} + a_{2} + ... + a_{12}\ \ \ \ \ \ \ \ \ \ \ \ \$….(ii)
On putting $x = - 1$ in Eq. (i), we get
$(1 - 1 - 2)^{6} = 1 - a_{1} + a_{2} - a_{3} + \ldots + a_{12}$
$\Longrightarrow {( - 2)}^{6} = 1 + a_{1} + a_{2} - a_{3} + ... + a_{12}$ …(iii)
On adding Eqs. (ii) and (iii) we get
$( - 2)^{6} = 2\left( 1 + a_{2} + a_{4} + \ldots + a_{12} \right)$
$\Longrightarrow \frac{64}{2} - 1 = a_{2} + a_{4} + \ldots + a_{12}$
$\therefore a_{2} + a_{4} + \ldots + a_{12} = 31$