Question

The sum of $\sum _{r=0}^n(-1{)}^r\frac{{}^n{C}_r}{{}^{r+2}{C}_r}$ is

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

Answer 1

The correct option is D $\frac{2}{n+2}$

We have,
$\sum _{r=0}^n(-1{)}^r\frac{{}^n{C}_r}{{}^{r+2}{C}_r}$

$=\sum _{r=0}^n(-1{)}^r\frac{n!}{(n-r)!r!}\times \frac{2!r!}{(r+2)!}$

$=2\sum _{r=0}^n(-1{)}^r\frac{n!}{(n-r)!(r+2)!}$

$=\frac{2}{(n+1)(n+2)}\sum _{r=0}^n(-1{)}^r\frac{(n+2)!}{\left\{\right(n+2)-(r+2\left)\right\}!(r+2)!}$

$=\frac{2}{(n+1)(n+2)}\sum _{r=0}^n(-1{)}^{r+2}{}^{n+2}{C}_{r+2}$

$=\frac{2}{(n+1)(n+2)}\sum _{s=2}^{n+2}(-1{)}^s{}^{n+2}{C}_s$

$=\frac{2}{(n+1)(n+2)}[\left(\sum _{s=0}^{n+2}(-1{)}^s{}^{n+2}{C}_s\right)-({}^{n+2}{C}_0-{}^{n+2}{C}_1)]$
$=\frac{2}{n+2}$