Question

If ${(1+x)}^n=\sum _{r=0}^n{C}_r{x}^r$ , then the value of the expression $C_0-\frac{C_1}{2}+\frac{C_2}{3}-\frac{C_3}{4}+\cdots +{(-1)}^n\frac{C_n}{n+1}$ equals

• A. $0$
• B. $\frac{1}{n+1}$
• C. $-\frac{1}{n+1}$
• D. None of these

Answer 1

The correct option is B $\frac{1}{n+1}$

$(1-x{)}^n=1-{}^n{C}_{1}x+{}^n{C}_2{x}^2+...(-1{)}^n{}^n{C}_n{x}^n$
Integrating both sides with respect to x, we get
$\frac{1}{n+1}[-(1-x{)}^{n+1}+1]=x-\frac{{}^n{C}_1{x}^2}{2}+\frac{{}^n{C}_2{x}^3}{3}+...(-1{)}^n{}^n{C}_n\frac{x^{n+1}}{n+1}$
Where $\frac{1}{n+1}$ is constant of integration.
Substituting $x=1$, we get
$\frac{1}{n+1}=1-\frac{{}^n{C}_1}{2}+\frac{{}^n{C}_2}{3}+...(-1{)}^n{}^n{C}_n\frac{1}{n+1}$