Question

The value of ${\sum }_{r=0}^n(-1{)}^r\left(\frac{{}^n{C}_r}{{}^{r+3}{C}_r}\right)$ is

• A. \N
• B. $\frac{3!}{2(n+3)}$
• C. $n$
• D. $3!\left(n^2\right)$

Answer 1

The correct option is B $\frac{3!}{2(n+3)}$

${\sum }_{r=0}^n(-1{)}^r\frac{{}^n{C}_r}{{}^{r+3}{C}_r}={\sum }_{r=0}^n(-1{)}^r\frac{n!.3!}{(n-r)!(r+3)!}=3!{\sum }_{r=0}^n(-1{)}^r\frac{n!}{(n-r)!(r+3)!}=\frac{3!}{(n+1)(n+2)(n+3)}{\sum }_{r=0}^n\frac{(-1{)}^r.(n+3)!}{(n-r)!(r+3)!}=\frac{3!}{(n+1)(n+2)(n+3)}{\sum }_{r=0}^n(-1{)}^r{}^{n+3}{C}_{r+3}=\frac{3!(-1{)}^3}{(n+1)(n+2)(n+3)}{\sum }_{s=3}^{n+3}(-1{)}^s{}^{n+3}{C}_s=\frac{-3!}{(n+1)(n+2)(n+3)}\left({\sum }_{s=0}^{n+3}\right(-1{)}^s{.}^{n+3}{C}_s)+\frac{-3!}{(n+1)(n+2)(n+3)}({-}^{n+3}{C}_0{+}^{n+3}{C}_1{-}^{n+3}{C}_2)=\frac{-3!}{(n+1)(n+2)(n+3)}(0-1+(n+3)-\frac{(n+3)(n+2)}{2!})=\frac{-3!}{(n+1)(n+2)(n+3)}.\frac{(n+2)(2-n-3)}{2}=\frac{3!}{2(n+3)}$