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

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(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} = $

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