If $(1 + x)^n = c_0 + c_1x + c_2x^2 + c_3x^3 + \dots + c_nx^n$,then the value of $c_0 - 3c_1 + 5c_2 - \dots + (-1)^n(2n + 1)c_n$ is

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

Explore More

Similar Questions

If $c_0, c_1, c_2, \ldots, c_n$ denote the coefficients in the expansion of $(1+x)^n$, then the value of $c_1 + 2c_2 + 3c_3 + \ldots + nc_n$ is

Evaluate the sum: $\left( \binom{21}{1} - \binom{10}{1} \right) + \left( \binom{21}{2} - \binom{10}{2} \right) + \left( \binom{21}{3} - \binom{10}{3} \right) + \dots + \left( \binom{21}{10} - \binom{10}{10} \right) = $

If $\binom{10}{2} + \binom{10}{3} + \binom{11}{4} + \binom{12}{5} + \binom{13}{6} = \binom{14}{r}$,then $r = \dots$

$\binom{50}{4} + \sum_{i=1}^{6} \binom{56-i}{3} = \dots$

If $\sum_{r=0}^{20} {}^{20+r}C_r = \frac{p}{q} {}^{40}C_{20}$ and $GCD(p, q) = 1$,then $p^2 - q^2 =$

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