Question

If $n$ is odd natural number, then ${\sum }_{r=0}^n\frac{{(-1)}^r}{{{}^{n}C}_r}$ equals................

Answer 1

$\sum _{r=0}^n\frac{{(-1)}^r}{{{}^{n}C}_r}=\frac{1}{{{}^{n}C}_0}-\frac{1}{{{}^{n}C}_1}+\frac{1}{{{}^{n}C}_2}+.......+\frac{1}{{{}^{n}C}_{n-1}}-\frac{1}{{{}^{n}C}_n}$ (Since $n$ is odd natural number)
$\sum _{r=0}^n\frac{{(-1)}^r}{{{}^{n}C}_r}=\frac{1}{{{}^{n}C}_n}-\frac{1}{{{}^{n}C}_{n-1}}+\frac{1}{{{}^{n}C}_{n-2}}+.......+\frac{1}{{{}^{n}C}_{n-1}}-\frac{1}{{{}^{n}C}_n}$ ($\because {{}^{n}C}_r={{}^{n}C}_{n-r}$ )
$=0$