Explanation

(a)
$\sum_{k = 0}^{10}{\ \ }\ ^{20}C_{k} = \ ^{20}C_{0} + \ ^{20}C_{1} + \ ^{20}C_{2} + \ldots + \ ^{20}C_{10}$
On putting $x = 1$and $n = 20$in $(1 + x)^{n}$
$\ \ \ \ \ \ \ \ \ = \ ^{n}C_{0} + \ ^{n}C_{1}x + \ ^{n}C_{2}x^{2} + \ldots + \ ^{n}C_{n}x^{n}$
We get
$\ 2^{20} = 2\left( \ ^{20}C_{0} + \ ^{20}C_{1} + \ ^{20}C_{2} + \ldots + \ ^{20}C_{9} \right) + \ ^{20}C_{10}$
$\Longrightarrow 2^{19} = \left( \ ^{20}C_{0} + \ ^{20}C_{1} + \ ^{20}C_{2} + \ldots + \ ^{20}C_{9} \right) + \frac{1}{2}\ ^{20}C_{10}$
$\Longrightarrow 2^{19} = \ ^{20}C_{0} + \ ^{20}C_{1} + \ ^{20}C_{2} + \ldots + \ ^{20}C_{10} - \frac{1}{2}\ ^{20}C_{10}$
$\Longrightarrow \ ^{20}C_{0} + \ ^{20}C_{1} + ... + \ ^{20}C_{10} = 2^{19} + \frac{1}{2}\ ^{20}C_{10}$