If $\sum_{k=1}^{30} k \left({ }^{30} C _k\right)^2 = \frac{\alpha 60 !}{(30 !)^2}$,then $\alpha$ is equal to

  • A
    $30$
  • B
    $60$
  • C
    $15$
  • D
    $10$

Explore More

Similar Questions

$\frac{C_0}{1} + \frac{C_1}{2} + \frac{C_2}{3} + .... + \frac{C_n}{n + 1} = $

Let $(1 + x)^m = C_0 + C_1x + C_2x^2 + C_3x^3 + . . . + C_mx^m$,where $C_r = {}^mC_r$ and $A = C_1C_3 + C_2C_4 + C_3C_5 + . . . + C_{m-2}C_m$. Which of the following is false?

Let $(1+x)^{10} = \sum_{r=0}^{10} c_{r} x^{r}$ and $(1+x)^{7} = \sum_{r=0}^{7} d_{r} x^{r}$. If $P = \sum_{r=0}^{5} c_{2r}$ and $Q = \sum_{r=0}^{3} d_{2r+1}$, then $\frac{P}{Q}$ is equal to:

The sum of the series $\binom{20}{0} - \binom{20}{1} + \binom{20}{2} - \binom{20}{3} + \dots + \binom{20}{10}$ is:

Let $n \in N$ and $[x]$ denote the greatest integer less than or equal to $x$. If the sum of $(n+1)$ terms ${}^{n}C_{0}, 3 \cdot {}^{n}C_{1}, 5 \cdot {}^{n}C_{2}, 7 \cdot {}^{n}C_{3}, \ldots$ is equal to $2^{100} \cdot 101$,then $2\left[\frac{n-1}{2}\right]$ is equal to $....$

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