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. none of these

Answer 1

Answer: (C)

$\frac{1}{3(n+1)}$

Given

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

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

We konw that

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

Intergrate w.r. to $x$ with limits from $-1to0$

${\left[\frac{(1+x{)}^{n+1}}{n+1}\right]}_{-1}^0={[C_{0}x+\frac{C_1{x}^2}{2}+\frac{C_2{x}^3}{3}+..............+\frac{C_n{x}^n+1}{n+1}]}_{-1}^0$

When putting limit from $-1to0$

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

Hence

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

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