Question

$C_0-\frac{C_1}{2}+\frac{C_2}{3}-\frac{C_3}{4}+...+{(-1)}^n\frac{C_n}{n+1}=\frac{1}{n+1}$

• A. True
• B. False

Answer 1

The correct option is A True

II Method: By Integration
${(1+x)}^n=C_0+C_{1}x+C_2{x}^2+...+C_n{x}^n$
Integrating both sides, within the limits $-1$ to $0.$
${\left[\frac{{(1+x)}^{n+1}}{n+1}\right]}_{-1}^0={[C_{0}x+C_1\frac{x^2}{2}+C_2\frac{x^3}{3}+...+C_n\frac{x^{n+1}}{n+1}]}_{-1}^0$
$\frac{1}{n+1}-0=0-[-C_0+\frac{C_1}{2}+\frac{C_2}{3}+...+{(-1)}^{n+1}\frac{C_n}{n+1}]$
$C_0-\frac{C_1}{2}+\frac{C_2}{3}-...+{(-1)}^n\frac{C_n}{n+1}=\frac{1}{n+1}$