Question

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

• A. $0$
• B. ${(-1)}^{\frac{n}{2}}(n+1)$
• C. ${(-1)}^{\frac{n}{2}}(n+2)$
• D. ${(-1)}^{\frac{n}{2}}n$

Answer 1

Answer: (C)

${(-1)}^{\frac{n}{2}}(n+2)$

$C_0^2-2C_1^2+3C_2^2-4C_3^2+....+{(-1)}^n(n+1)C_n^2$

$=(C_0^2-C_1^2+C_2^2-C_3^2+....+{(-1)}^n{C}_n^2)-(C_1^2-2C_2^2+3C_3^2+....+{(-1)}^n{nC}_n^2)$

$={(-1)}^{\frac{n}{2}}\frac{n!}{\left(\frac{n}{2}\right)!\left(\frac{n}{2}\right)!}-{(-1)}^{\frac{n}{2}-1}\frac{n}{2}{{}^{n}C}_{\frac{n}{2}}={(-1)}^{\frac{n}{2}}\frac{n!}{\left(\frac{n}{2}\right)!\left(\frac{n}{2}\right)!}(1+\frac{n}{2})$

Thus $2\left(\frac{n}{2}\right)!\left(\frac{n}{2}\right)![C_0^2-{2C}_1^2+{3C}_2^2-C_3^2+....+{(-1)}^n(n+1)C_n^2]={(-1)}^{\frac{n}{2}}(n+2)$