Question

If n is a positive integer then prove that
$C_0+\frac{C_1}{2}+\frac{C_2}{3}+..........+\frac{C_n}{n+1}=\frac{2^{n+1}-1}{n+1}$

Answer 1

${\left(1+x\right)}^2=C_0+C_{1}x+C_2{x}^2+......C_n{x}^{27}$

Integratinmg both sides from $0$ to $1$

$\begin{array}{c}{\int }_0^1{\left(1+x\right)}^{n}dx={\left[C_{0}x+\frac{C_1{x}^2}{2}+\frac{C_2{x}^3}{3}+....+\frac{C_n{x}^{n+1}}{\left(n+1\right)}\right]}_0^1\\ {\left[\frac{{\left(1+x\right)}^{n+1}}{\left(n+1\right)}\right]}_0^1=C_0+\frac{C_1}{2}+\frac{C_2}{3}+...+\frac{C_n}{n+1}\\ C_0+\frac{C_1}{2}+\frac{C_2}{3}+....+\frac{C_n}{n+1}=\frac{2^{n+1}-1}{\left(n+1\right)}\end{array}$

Hence, proved.