Question

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

Answer 1

You may proceed exactly as in part (a)
Alternate method is given below:
$\frac{{}^n{C}_r}{{}^{r+2}{C}_r}=\frac{n!}{(n-r)!r!}\frac{(r+2)!}{2!r!}=\frac{2\left(n\right)!}{(n-r)!(r+2)!}$
$=\frac{2}{(n+2)(n+1)}\left[\frac{(n+2)!}{\left[\right(n+2)-(r+2)!](r+2)!}\right]$
$=\frac{2}{(n+2)(n+1)}{}^{n+2}{C}_{r+2}$ .....(A)
Now put r + 2 = s, when r = 0, s= 2, when r = n, s = n + 2
$\therefore ,\sum =\frac{2}{(n+2)(n+1)}{\sum }_{s=2}^{n+2}(-1{)}^{s+2}\cdot {}^{n+2}{C}_s$
$=\frac{2}{(n+2)(n+1)}$
$\left[{\sum }_{s=2}^{n+2}\right(-1{)}^s{}^{n+2}{C}_s-{[}^{n+2}{C}_0{-}^{n+2}{C}_1\left]\right]$
$=\frac{2}{(n+2)(n+1)}[0-[1-(n+2)\left]\right]$
$=\frac{2}{(n+2)(n+1)}(n+1)=\frac{2}{(n+2)}$