If $C_{r}$ stands for $^{n} C_{r^{'}}$ then the sum of the series
$\frac{2 (\begin{array}{l} n \\ 2 \end{array})! (\begin{array}{l} n \\ 2 \end{array})!}{n!} [C_{0}^{2} - 2 C_{1}^{2} + 3 C_{2}^{2} + \dots + (- 1)^{n} (n + 1) C_{n}^{2}]$
where $n$ is an even positive integer, is equal to

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$(- 1)^{n / 2} (n + 1)$

$(- 1)^{n / 2} (n + 2)$

$(- 1)^{n} n$

answer is C.

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Detailed Solution

We have

$\begin{aligned} C_{0}^{2} - 2 C_{1}^{2} + 3 C_{2}^{2} - 4 C_{3}^{2} + \dots + (- 1)^{n} (n + 1) C_{n}^{2} \\ = \{C_{0}^{2} - C_{1}^{2} + C_{2}^{2} - C_{3}^{2} + \dots + (- 1)^{n} C_{n}^{2}\} \end{aligned}$

$\{- C_{1}^{2} - 2 C_{2}^{2} + 3 C_{3}^{2} - \dots + (- 1)^{n} n C_{n} 2$

$= (- 1)^{n / 2} \frac{n!}{(\frac{n}{2})! (\begin{array}{l} n \\ 2 \end{array})!} - (- 1)^{\frac{n}{2}} - {\frac{1}{2}}^{n} C_{n}^{n}$

$= (- 1)^{n / 2} \frac{n!}{(\begin{array}{l} n \\ 2 \end{array})! (\begin{array}{l} n \\ 2 \end{array})!} (1 + \frac{n}{2})$

$\begin{array}{l} ∴ 2 (\frac{n}{2})! (\frac{n}{2})! \{C_{0}^{2} - 2 C_{1}^{2} - \dots + (- 1)^{n} (n + 1) {C_{n}}^{2}\} \\ = (- 1)^{n / 2} (n + 2) . \end{array}$