Explanation

(c)
Since, ${(1 + x - 3x}^{2})^{10} = 1 + a_{1}x + a_{2}x^{2} + \ldots + a_{20}x^{20}$
On putting $x = - 1,$ we get
$(1 - 1 - 3)^{10} = 1 - a_{1} + a_{2} - \ldots + a_{20} = 3^{10}\ldots(i)$
Again putting $x = 1$, we get
$(1 + 1 - 3)^{10} = 1 + a_{1} + a_{2} - \ldots + a_{20} = 1\ \ \ldots(ii)$
On adding Eqs. (i) and (ii), we get
$2\left( 1 + a_{2} + a_{4} + \ldots + a_{20} \right) = 3^{10} + 1$
$\Longrightarrow a_{2} + a_{4} + ... + a_{20} = \frac{3^{10} + 1}{2} - 1 = \frac{3^{10} - 1}{2}$