Question

$\frac{C_1}{2}+\frac{C_3}{4}+\frac{C_5}{6}+...=\frac{2^n-1}{n+1}$

Answer 1

The $r_{th}$ term in the given expansion is

$T_r=\frac{{}^n{C}_{2r-1}}{2r}$

Since $\frac{1}{r+1}\cdot {}^n{C}_r=\frac{1}{n+1}\cdot {}^{n+1}{C}_{r+1}$

$\therefore T_r=\frac{{}^n{C}_{2r-1}}{2r}=\frac{1}{n+1}\cdot {}^{n+1}{C}_{2r}$

$\therefore \frac{{}^n{C}_1}{2}+\frac{{}^n{C}_3}{4}+\frac{{}^n{C}_5}{6}+......=\frac{1}{n+1}[{}^{n+1}{C}_2+{}^{n+1}{C}_4+.....]$

$=\frac{1}{n+1}[2^{n+1-1}-{}^n{C}_0]=\frac{2^n-1}{n+1}$