Question

If $n\in N$, then the value of
$S={\sum }_{r=0}^n{(-1)}^r\frac{\left({{}^{n}C}_r\right)}{\left({{}^{r+2}C}_r\right)}$ is

• A. $\frac{1}{n+2}$
• B. $\frac{2}{n+2}$
• C. $n+2$
• D. $n+1$

Answer 1

Answer: (B)

$\frac{2}{n+2}$

$\frac{\left({{}^{n}C}_r\right)}{\left({{}^{r+2}C}_r\right)}=\frac{n!}{(n-r)!r!}\cdot \frac{r!2!}{(r+2)!}$
$=\frac{2!}{(n+1)(n+2)}\cdot \frac{(n+2)!}{(r+2)!(n-r)!}$
$=\frac{2!}{(n+1)(n+2)}\left({{}^{n+2}C}_{r+2}\right)$
$\therefore S=\frac{2}{(n+1)(n+2)}{\sum }_{r=0}^n{(-1)}^{r+2}\left({{}^{n+2}C}_{r+2}\right)$
$=\frac{2}{(n+1)(n+2)}[{\sum }_{r=0}^{n+2}{(-1)}^{r+2}\left({{}^{n+2}C}_r\right)-{{}^{n+2}C}_0+{{}^{n+2}C}_{10}]$
$=\frac{2}{(n+1)(n+2)}[0+(n+1\left)\right]=\frac{2}{n+2}$