Question

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

• A. $\frac{2^n}{n+1}$
• B. $\frac{2^n-1}{n+1}$
• C. $\frac{2^{n+1}-1}{n+1}$
• D. $Noneofthese$

Answer 1

The correct option is A $\frac{2^n}{n+1}$

Solve:

$(1+x{)}^n=c_0{x}^0+c_{1}x+c_2{x}^2+\dots =c_n{x}^n$

On integrating both sides we get,

$\frac{(1+x{)}^{n+1}}{n+1}=c_{0}x+c_1{x}_2^2+c_2{x}_3^3+\cdots \cdot \frac{c_n{x}^{n+1}}{n+1}$

on Putting $x=1$ we get,

$\frac{2^{n+1}}{n+1}=c_0+\frac{c_1}{2}+\frac{c_2}{3}+\cdots \frac{c_n}{n+1}$

$\Rightarrow \frac{c_0}{1}+\frac{c_1}{2}+\frac{c_2}{3}+\cdots -\frac{c_n}{n+1}=\frac{2^{n+1}}{n+1}$