Explanation

(c)
Given, ${(1 + 2x + x}^{2})^{5} = \sum_{k = 0}^{15}{\ \ }a_{k}x^{k}$
$\Longrightarrow (1 + x)^{10} = a_{0}x^{0} + a_{1}x + a_{2}x^{2} + \ldots + a_{15}x^{15}$
$\Longrightarrow \ ^{10}C_{0} + \ ^{10}C_{1}x + \ ^{10}C_{2}x^{2} + \ldots + \ ^{10}C_{10}x^{10}$
$\ \ \ \ \ \ \ \ \ \ = a_{0} + a_{1}x + a_{2}x^{2} + a_{3}x^{3} + \ldots + a_{15}x^{15}$
On equating the coefficient of constant and even power of $x$, we get
$\ \ \ \ \ \ \ \ \ \ \ \ a_{0} = 10_{C_{0}},\ a_{2} = 10_{C_{2}},$
$\ \ \ \ a_{4} = \ ^{10}C_{4},\$…..$a_{10} = \ ^{10}C_{10},\ a_{12} = a_{14} = 0$
$\therefore\ \sum_{k = 0}^{7}{\ \ }a_{2k} = \ ^{10}C_{0} + \ ^{10}C_{2} + \ ^{10}C_{4} + \ ^{10}C_{6}$
+$\ ^{10}C_{8} + \ ^{10}C_{10} + 0 + 0$
$= 2^{10 - 1} = 2^{9} = 512$