Question

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

Answer 1

LHS$=\frac{C_1}{2}+\frac{C_3}{4}+\frac{C_5}{6}+....$

Case$1$If $n$ is odd say $n=2m+1$ for all $m\in W$ then

LHS$={\sum }_{r=0}^m\frac{{{2m+1}_C}_{2r+1}}{2r+2}$

$={\sum }_{r=0}^m\frac{{{2m+2}_C}_{2r+2}}{2m+2}$ $(\because \frac{{{2m+2}_C}_{2r+2}}{2m+2}=\frac{{{2m+1}_C}_{2r+2}}{2r+2})$

$=\frac{1}{(2m+2)}({{2m+2}_C}_2+{{2m+2}_C}_4+{{2m+2}_C}_6+...+{{2m+2}_C}_{2m+2})$

$=\frac{1}{(2m+2)}(2^{2m+2-1}-{{2m+2}_C}_0)$

$=\frac{2^n-1}{n+1}$ by taking $2m+1=n$

$=$RHS

Case$2$If $n$ is even say $n=2m$ for all $m\in N$ then

LHS$={\sum }_{r=0}^m\frac{{{2m}_C}_{2r+1}}{2r+2}$

$={\sum }_{r=0}^m\frac{{{2m+1}_C}_{2r+2}}{2m+1}$ $(\because \frac{{{2m+1}_C}_{2r+2}}{2m+1}=\frac{{{2m+1}_C}_{2r+1}}{2r+2})$

$=\frac{1}{(2m+1)}({{2m+1}_C}_2+{{2m+1}_C}_4+{{2m+1}_C}_6+...+{{2m+2}_C}_{2m})$

$=\frac{1}{(2m+1)}(2^{2m+1-1}-{{2m+2}_C}_0)$

$=\frac{2^n-1}{n+1}$ by taking $2m=n$

$=$RHS