Question

For $n\in N$. let
$S\left(n\right)={\sum }_{r=0}^n{(-1)}^r\frac{1}{\left({{}^{n}C}_r\right)}$
Value of $S={\sum }_{r=0}^n{(-1)}^r\frac{\left({{}^{r+2}C}_r\right)}{\left({{}^{n}C}_r\right)}$ is

• A. $S(n+2)-\frac{1}{2}{(n+1)}^2$
• B. $S(n+2)+\frac{1}{2}{(n+1)}^2$
• C. $0$
• D. none of these

Answer 1

The correct option is D none of these

$\frac{\left({{}^{r+2}C}_r\right)}{\left({{}^{n}C}_r\right)}=\frac{(n+1)(n+2)}{2}\frac{1}{{{}^{n+2}C}_{r+2}}$
Thus, $S=\frac{1}{2}(n+1)(n+2)\sum _{r=0}^n{(-1)}^r\frac{1}{{{}^{n+2}C}_{r+2}}$
$=\frac{1}{2}(n+1)(n+2)[\sum _{r=0}^n{(-1)}^r\frac{1}{{{}^{n+2}C}_r}-\frac{1}{{{}^{n+2}C}_0}+\frac{1}{{{}^{n+2}C}_1}]$
$=\frac{1}{2}(n+1)(n+2)\left[S\right(n+2)-1+\frac{1}{n+2}]$
$=\frac{1}{2}(n+1)(n+2)S(n+2)-\frac{1}{2}{(n+1)}^2$