If $C_r$ stands for ${}^{n}C_r$, then the sum of the series $\frac{2\left(\frac{n}{2}\right)!\left(\frac{n}{2}\right)!}{n!}\left[C_0^2-2C_1^2+3C_2^2\mathrm{.......}+{\left(-1\right)}^n\left(n+1\right)C_n^2\right]$ where ‘n’ is an even positive integer is Equal to
 

$n\text{ is even so n = 2m}$

$E=2.\frac{m!\text{ m!}}{\left(\text{2m}\right)!}.\left[C_0^2-2C_1^2+3C_2^2\mathrm{......}+{\left(-1\right)}^{2m}\left(2m+1\right)C_{2m}^2\right]\to \left(1\right)$

$C_{2m}=C_0,{\text{ C}}_{\text{2m-1}}=C_1$

Write the terms in reverse order

$E=\frac{2m!m!}{\left(2m\right)!}\left[\left(2m+1\right)C_0^2-2m{C}_1^2+\left(2m-1\right)C_2^2+\mathrm{.....}{C}_{2m}^2\right]\to \left(2\right)$

$2E=\frac{2m!m!}{\left(2m\right)!}\left(2m+2\right)\left(C_0^2-C_1^2+C_2^2+\mathrm{.....}{C}_{2m}^2\right)$

$=\frac{2m!m!\left(m+1\right)}{\left(2m\right)!}{\left(-1\right)}^{m}2m{C}_m\left[C_0^2-C_1^2+C_2^2\mathrm{....}+C_n^2\right]={\left(-1\right)}^{n/2}.n{C}_{n/2}$

$=2{\left(-1\right)}^m\left(m+1\right)=2{\left(-1\right)}^{n/2}\left(\frac{n}{2}+1\right)={\left(-1\right)}^{n/2}\left(n+2\right)$