Question

If ${(1+x)}^n={\sum }_{r=0}^n{C}_r{x}^r$, show that
$C_0+\frac{C_1}{2}+....+\frac{C_n}{n+1}=\frac{2^{n+1}-1}{n+1}$

Answer 1

$(1+x{)}^n={\Sigma }_{r=0}^n{C}_r{x}^r$

Integrating on both sides

${\int }_0^1(1+x{)}^{n}dx={\Sigma }_{r=0}^n{C}_r{\int }_0^1{x}^{r}dx$

$[\frac{(1+x{)}^{n+1}}{n+1}{]}_0^1=[{\Sigma }_{r=0}^n{C}_r\frac{x^{r+1}}{r+1}{]}_0^1$

$\frac{2^{n+1}-1}{n+1}={\Sigma }_{r=0}^n\frac{C_r}{r+1}$

$⟹C_0+\frac{C_1}{2}+\cdots +\frac{C_n}{n+1}=\frac{2^{n+1}-1}{n+1}$