$C_0 - C_1 + C_2 - C_3 + \dots + (-1)^n C_n$ is equal to
- A$2^n$
- B$2^n - 1$
- C$0$
- D$2^{n-1}$
Explore More
Similar Questions
If $^nC_r = C_r$ and $2 \frac{C_1}{C_0} + 4 \frac{C_2}{C_1} + 6 \frac{C_3}{C_2} + \dots + 2n \frac{C_n}{C_{n-1}} = 650$,then $^nC_2 =$
If $C_r = { }^n C_r$,then find the sum $C_0 + C_4 + C_8 + C_{12} + \ldots$
DifficultAP EAMCET 2018
View SolutionThe value of $\frac{^{100}C_{50}}{51} + \frac{^{100}C_{51}}{52} + \dots + \frac{^{100}C_{100}}{101}$ is:
DifficultJEE MAIN 2026
View SolutionIf $n$ is a positive integer greater than $1$,then $3({ }^n C_0) - 8({ }^n C_1) + 13({ }^n C_2) - 18({ }^n C_3) + \ldots$ up to $(n+1)$ terms $=$
If $(1 + x)^n = C_0 + C_1x + C_2x^2 + ... + C_nx^n$,then the value of $C_0 + C_2 + C_4 + C_6 + ...$ is
Easy
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo