Explanation

(d)
We have, $C_{0}^{2} - 2C_{1}^{2} + 3C_{2}^{2} - ... + ( - 1)^{n}(n + 1)C_{n}^{2}$
$= \left\lbrack C_{0}^{2} - C_{1}^{2} + C_{2}^{2} - ... + {( - 1)}^{n}C_{n}^{2} \right\rbrack - \left\lbrack C_{1}^{2} - 2C_{2}^{2} + 3C_{3}^{2} - ... + {( - 1)}^{n}nC_{n}^{2} \right\rbrack$
$= {( - 1)}^{\frac{n}{2}}\frac{n!}{\left( \frac{n}{2} \right)!\left( \frac{n}{2} \right)!} \bullet - {( - 1)}^{\frac{n}{2 - 1}} \bullet \frac{1}{2}n\frac{n!}{\left( \frac{n}{2} \right)!\left( \frac{n}{2} \right)!}$
$= {( - 1)}^{\frac{n}{2}} \bullet \frac{n!}{\left( \frac{n}{2} \right)!\left( \frac{n}{2} \right)!} \bullet \left( 1 + \frac{n}{2} \right)$
Therefore, the value of the given expression is
$\frac{2\left( \frac{n}{2} \right)!\left( \frac{n}{2} \right)!}{n!} \times {( - 1)}^{\frac{n}{2}} \bullet \frac{(n)!}{\left( \frac{n}{2} \right)!\left( \frac{n}{2} \right)!}\left( 1 + \frac{n}{2} \right) = ( - 1)^{\frac{n}{2}}(2 + n)$