^nC_0 - frac{1}{2} ^nC_1 + frac{1}{3} ^nC_2 - | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) We know that $(1-x)^n = {^nC_0} - {^nC_1} x + {^nC_2} x^2 - \dots + (-1)^n {^nC_n} x^n$.
Integrating both sides with respect to $x$ from $0$ to $

Vedclass · Accepted Answer

$\frac{1}{n+1}$

Vedclass · Answer

(C) We know that $(1-x)^n = {^nC_0} - {^nC_1} x + {^nC_2} x^2 - \dots + (-1)^n {^nC_n} x^n$.
Integrating both sides with respect to $x$ from $0$ to $1$:
$\int_0^1 (1-x)^n \, dx = \int_0^1 ({^nC_0} - {^nC_1} x + {^nC_2} x^2 - \dots + (-1)^n {^nC_n} x^n) \, dx$.
Evaluating the integral on the left side:
$\left[ -\frac{(1-x)^{n+1}}{n+1} \right]_0^1 = 0 - \left( -\frac{1}{n+1} \right) = \frac{1}{n+1}$.
Evaluating the integral on the right side:
$^nC_0 [x]_0^1 - {^nC_1} \left[ \frac{x^2}{2} \right]_0^1 + {^nC_2} \left[ \frac{x^3}{3} \right]_0^1 - \dots + (-1)^n {^nC_n} \left[ \frac{x^{n+1}}{n+1} \right]_0^1 = {^nC_0} - \frac{1}{2} {^nC_1} + \frac{1}{3} {^nC_2} - \dots + \frac{(-1)^n {^nC_n}}{n+1}$.
Thus, the sum is equal to $\frac{1}{n+1}$.

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

Similar Questions

If $(1 + x)^n = C_0 + C_1x + C_2x^2 + .... + C_nx^n$,then $C_0C_2 + C_1C_3 + C_2C_4 + .... + C_{n-2}C_n$ equals

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

If $p$ and $q$ are positive integers,then the coefficients of $x^p$ and $x^q$ in the expansion of $(1 + x)^{p + q}$ are

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

The value of $^{15}C_0^2 - ^{15}C_1^2 + ^{15}C_2^2 - ... - ^{15}C_{15}^2$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module