Explanation

(a)
${(1 + x)}^{n} = \ ^{n}C_{0} + \ ^{n}C_{1}x^{\ } + \ ^{n}C_{2}x^{2} + ... + \ ^{n}C_{n}x^{n}$ …(i)
On differentiating both sides w. r. t. $x,$ we get
$n(1 + {x)}^{n - 1} = \ ^{n}C_{1} + 2\ \ ^{n}C_{2}x + ... + n\ ^{n}C_{n}x^{n - 1}$ …(ii)
On putting $x = 1$ in Eq. (ii), we get
${n(2)}^{n - 1} = \ ^{n}C_{1} + 2\ \ ^{n}C_{2} + ... + n\ \ ^{n}C_{n}$ …(iii)
On putting $x = - 1$ in Eq. (ii) we get
$0 = \ ^{n}C_{1} - 2\ ^{n}C_{2} + 3\ \ ^{n}C_{3} - ...( - 1)^{n - 1} \bullet \ ^{n}C_{n}\ldots$(iv)
On adding Eqs. (iii) and (iv), we get
$n2^{n - 1} = 2\left( \ ^{n}C_{1} + 3\ \ ^{n}C_{3} + .. \right)$
$\Longrightarrow \ ^{n}C_{1} + 3\ \ ^{n}C_{3} + 5\ \ ^{n}C_{5} + ... = \frac{n}{2} \bullet 2^{n - 1} = n2^{n - 2}$