Question

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

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

Answer 1

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

$\sum _{r=0}^n(-1{)}^r\frac{{}^n{C}_r}{{}^{r+2}{C}_r}=\sum _{r=0}^n(-1{)}^r\frac{n!}{(n-r)!r!}\cdot \frac{r!\cdot 2!}{(r+2)!}$
$=2\sum _{r=0}^n(-1{)}^r\frac{n!}{(n-r)!(r+2)!}$
$=\frac{2}{(n+1)(n+2)}\sum _{r=0}^n(-1{)}^r\frac{(n+2)!}{\left[\right(n+2)-(r+2\left)\right]!(r+2)!}$
$=\frac{2}{(n+1)(n+2)}\sum _{r=0}^n(-1{)}^r{}^{n+2}{C}_{r+2}$
$=\frac{2}{(n+1)(n+2)}({}^{n+2}{C}_2-{}^{n+2}{C}_3+{}^{n+2}{C}_4-\cdots +(-1{)}^n{}^{n+2}{C}_{n+2})$

Add and subtract ${}^{n+2}{C}_0-{}^{n+2}{C}_1,$ we get
$\sum _{r=0}^n(-1{)}^r\frac{{}^n{C}_r}{{}^{r+2}{C}_r}$
$=\frac{2}{(n+1)(n+2)}[({}^{n+2}{C}_0-{}^{n+2}{C}_1+{}^{n+2}{C}_2-\cdots +(-1{)}^n{}^{n+2}{C}_{n+2})-{(}^{n+2}{C}_0-{}^{n+2}{C}_1)]$
$=\frac{2}{(n+1)(n+2)}[0-(1-(n+2)\left)\right]$
$=\frac{2}{n+2}$