Question

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

• A. $0$
• B. ${(-1)}^{\frac{n}{2}}(n+1)$
• C. ${(-1)}^n(n+2)$
• D. ${(-1)}^{n}n$
• E. none

Answer 1

Answer: (E)

none

$C_0^2-{2C}_1^2+{3C}_2^2-{4C}_3^2+...+{(-1)}^n(n+1)C_n^2=\{C_0^2-C_1^2+C_2^2-C_3^2+...+{{(-1)}^{n}C}_n^2\}-\{C_1^2-{2C}_2^2+3C_3^2-...+{(-1)}^n{nC}_n^2\}$
$={(-1)}^{\frac{n}{2}}\frac{n!}{\left(\frac{n}{2}\right)!\left(\frac{n}{2}\right)!}-{(-1)}^{\frac{n}{2}-1}\frac{n}{2}\frac{n!}{\left(\frac{n}{2}\right)!\left(\frac{n}{2}\right)!}$
$={(-1)}^{\frac{n}{2}}\frac{n!}{\left(\frac{n}{2}\right)!\left(\frac{n}{2}\right)!}(1+\frac{n}{2})$
$\therefore \frac{2\left(\frac{n}{2}\right)!\left(\frac{n}{2}\right)!}{n!}\{C_0^2-{2C}_1^2+{3C}_2^2-...+{(-1)}^r(n+r)C_n^2\}$
$=\frac{2\left(\frac{n}{2}\right)!\left(\frac{n}{2}\right)!}{n!}{(-1)}^{\frac{n}{2}}\frac{n!}{\left(\frac{n}{2}\right)!\left(\frac{n}{2}\right)!}\frac{(n+2)}{2}={(-1)}^{\frac{n}{2}}(n+2)$