Question

Prove that `(3!)/(2(n+3)) =sum_(r=0)^(n) (-1)^(r ) ((""^(n)C_(r ))/(""^(r+3)C_(r )))`

Answer 1

`underset(r=0)overset(n)sum(-1)^(r ) ((.^(n)C_(r))/(.^(r+3)C_(3)))`
`= underset(r=0)overset(n)sum(-1)^(r)(n!)/((n-r)!r!)(3!r!)/((r+3)!)`
`= 3!underset(r=0)overset(n)sum(-1)^(r) (n!)/((n-r)!(r+3!))`
`=(3!)/((n+1)(n+2)(n+3))underset(r=0)overset(n)sum(-1)^(r).^(n+3)C_(r+3)`
` = - (3!)/((n+1)(n+2)(n+3))underset(r=0)overset(n)sum(-1)^(r+3).^(n+3)C_(r+3)`
`= - (3!)/((n+1)(n+2)(n+3))[-.^(n+3)C_(3) +.^(n+3)C_(4)-"....."+(-1)^(n+3).^(n+3)C_(n+3)]`
`= - (3!)/((n+1)(n+2)(n+3))[(.^(n+3)C_(0)-.^(n+3)C_(1)+.^(n+3)C_(2)-.^(n+3)C_(3)+"...."+(-1)^(n+3).^(n+3)C_(n+3))-(.^(n+3)C_(0)-.^(n+3)C_(1)+.^(n+3)C_(2))]`
`- (3!)/((n+1)(n+2)(n+3))[(1-1)^(n+3)-(1-(n+3))-((n+3)(n+2))/(2)]`
`= (3!)/((n+1)(n+2)(n+3))[1-n-3+((n+3)(n+2))/(2)]`
`= (3!)/((n+1)(n+2)(n+3)) ((n^(2)+3n+2))/(2)`
`= (3!)/(2(n+3))`