If $(1 + x)^n = C_0 + C_1x + C_2x^2 + .......... + C_nx^n$,then $\frac{C_1}{C_0} + \frac{2C_2}{C_1} + \frac{3C_3}{C_2} + .... + \frac{nC_n}{C_{n - 1}} = $

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

Explore More

Similar Questions

$^nC_0 - \frac{1}{2} ^nC_1 + \frac{1}{3} ^nC_2 - \dots + (-1)^n \frac{^nC_n}{n+1} = $

Difficult
View Solution

If $(1 + x)^{15} = C_0 + C_1x + C_2x^2 + ...... + C_{15}x^{15},$ then $C_2 + 2C_3 + 3C_4 + .... + 14C_{15} = $

The sum to $(n + 1)$ terms of the series $\frac{C_0}{2} - \frac{C_1}{3} + \frac{C_2}{4} - \frac{C_3}{5} + \dots$ is

Difficult
View Solution

$\sum\limits_{r = 0}^m {^{n + r}{C_n} = } $

Difficult
View Solution

If $(1+x+x^2)^n = a_0 + a_1 x + a_2 x^2 + \ldots + a_{2n} x^{2n}$,then $a_0 + a_2 + a_4 + \ldots + a_{2n} =$

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