`sum_(r=0)^(n)(-1)^(r)(""^(n)C_(r))/(""^(r+3)C_(r))` is equal to :

Options

A. `(1)/(n+1)`
B. `(1)/(n+2)`
C. `(2)/(n+1)`
D. `(2)/(n+2)`

Answer

d

Explanation

We have,
`sum_(r=0)^(n) (-1)^(r) (""^(n)C_(r))/(""^(r+ 2)C_(r))` ltbrge `sum_(r=0)^(n) (-1)^(r) (n!)/((n+r)r!) xx(2!r!)/((r+2)!)`
=`sum_(r=0)^(n) (-1)^(r) (n!)/((n-r)(r+2))`
= `(2)/((n+1)(r+2))sum_(r=0)^(n) (-1)^(r+2)((n+2)!)/((n-r)(r+2)!)`
= `(2)/((n+1)(r+2))sum_(r=0)^(n) ""^(n+2)C_(r+ 2)(-1)^(r+2)`
=`(2)/((n+1)(r+2)){sum_(r=0)^(n)(-1)^(r +2) ""^(n+2)C_(r+ 2)""^(n+2)C_(0)+""^(n+2)C_(1)}`
`(2)/((n+1)(r+2)){0-1+n+2)= (2)/(n+2)`