Question

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

• A. $\frac{1}{3(n+1)}$
• B. $\frac{1}{(n+1)}$
• C. $\frac{3}{(n+1)}$
• D. None of these

Answer 1

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

We hve. ${(1+x)}^n=C_0+C_{1}x+C_2{x}^2+....+C_n{x}^n$

Integrating both sides w.r.t $x$ from $-1$ to $0$, we get

${\int }_{-1}^0{(1+x)}^{n}dx={\int }_{-1}^0(C_0+C_{1}x+C_2{x}^2+....+C_n{x}^n)dx$

$\Rightarrow {\left[\frac{{(1+x)}^{n+1}}{n+1}\right]}_{-1}^0={\left[C_{0}x+\frac{C_1{x}^2}{2}+....+\frac{C_n{x}^{n+!}}{n+1}\right]}_{-1}^0$

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

$\therefore$ The given expression

$=\frac{1}{3}(C_0-\frac{C_1}{2}+\frac{C_2}{3}-...)=\frac{1}{3}\times \frac{1}{n+1}=\frac{1}{3(n+1)}$