Question

The sum 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

Answer: (C)

$\sum _{r=0}^n(-1{)}^r\frac{{}^n{C}_r}{{}^{r+2}{C}_r}=\sum _{r=0}^n(-1{)}^r\frac{n!}{(n-r)!r!}\cdot \frac{r!\cdot 2!}{(r+2)!}$
$=2\sum _{r=0}^n(-1{)}^r\frac{n!}{(n-r)!(r+2)!}=\frac{2}{(n+1)(n+2)}\cdot \sum _{r=0}^n(-1{)}^r\frac{(n+2)!}{(n+2)-(r+2)!(r+2)!}$
$=\frac{2}{(n+1)(n+2)}\cdot \sum _{r=0}^n(-1{)}^r\cdot {}^{n+2}{C}_{r+2}$
$=\frac{2}{(n+1)(n+2)}\cdot \left({}^{n+2}{C}_2-{}^{n+2}{C}_3+{}^{n+2}{C}_4-{}^{n+2}{C}_5-{}^{n+2}{C}_{n+2}\right)\text{ add and subtract }{}^{n+2}{C}_0-{}^{n+2}{C}_1$
$=\frac{2}{(n+1)(n+2)}\cdot \left\{\left({}^{n+2}{C}_0-{}^{n+2}{C}_1+{}^{n+2}{C}_2-{}^{n+2}{C}_3+{}^{n+2}{C}_4-{}^{n+2}{C}_5-{}^{n+2}{C}_{n+2}\right)-\left({}^{n+2}{C}_0-{}^{n+2}{C}_1\right)\right\}$
$=\frac{2}{(n+1)(n+2)}(0-(1-(n+2)))=\frac{2}{n+2}$