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

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

Explore More

Similar Questions

If $\frac{1}{n+1} {}^{n}C_{n} + \frac{1}{n} {}^{n}C_{n-1} + \dots + \frac{1}{2} {}^{n}C_{1} + {}^{n}C_{0} = \frac{1023}{10}$,then $n$ is equal to

The value of $\binom{30}{0}\binom{30}{10} - \binom{30}{1}\binom{30}{11} + \binom{30}{2}\binom{30}{12} - ....... + \binom{30}{20}\binom{30}{30}$ is:

If the coefficients of $a^{r-1}$,$a^{r}$,and $a^{r+1}$ in the expansion of $(1+a)^{n}$ are in arithmetic progression,prove that $n^{2}-n(4r+1)+4r^{2}-2=0$.

Difficult
View Solution

If $(1 + x)^n = C_0 + C_1x + C_2x^2 + .......... + C_nx^n,$ then $C_0^2 + C_1^2 + C_2^2 + C_3^2 + ...... + C_n^2$ =

Difficult
View Solution

If $n$ is a positive integer,then $\sum_{r=1}^n r^2 \cdot C_r = (\ldots \ldots \ldots) 2^{n-2}$

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo