Question

The value of $\frac{C_0}{1.3}-\frac{C_1}{2.3}+\frac{C_2}{3.3}-\frac{C_3}{4.3}+....+{(-1)}^n\frac{C_n}{(n+1).3}$is

• A. $\frac{3}{n+1}$
• B. $\frac{n+1}{3}$
• C. $\frac{1}{3(n+1)}$
• D. $\frac{3}{n-1}$

Answer 1

The correct option is C $\frac{1}{3(n+1)}$

Consider the expansion ${(1-x)}^n=C_0-C_{1}x+C_2{x}^2-...{(-1)}^n{x}^n$,

Now, integrating the expansion with respect to $x$ from $0$ to $1$ gives,

$\frac{1}{n+1}=\frac{C_0}{1}-\frac{C_1}{2}+\frac{C_2}{2}-...{(-1)}^n\frac{C_n}{n+1}⟹\frac{1}{3(n+1)}=\frac{C_0}{1.3}-\frac{C_1}{2.3}+\frac{C_2}{3.3}-...{(-1)}^n\frac{C_n}{(n+1).3}$