1.

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

Answer»

Solution :`sum_(R=0)^(n) (-1) (""^(n)C_(r))/(""^(r+3)C_(r))`
`= sum_(r=0)^(n) (-1)^(r)(n!.3!)/((n-r)!(r+3))=3!sum_(r=0)^(n) (-1)^(r)(n!)/((n-r)!(r+3)!)`
`= (3!)/((n+1)(n+2)(n+3)).sum_(r=0)^(n) ((-1)^(r).(n+3)!)/((n-r)!(r+3)!)`
`=(3!)/((n+1)(n+2)(n+3)) =.sum_(r=0)^(n) (-1)^(r).""^(n+3)C_(r+3)`
`= (3!(-1)^(3))/((n+1)(n+2)(n+3))sum_(s=0)^(n) (-1)^(s).""^(n+3)C_(3)`
`=(-3!)/((n+1)(n+2)(n+3))(sum_(s=0)^(n+3) (-1)^(s).""^(n+3)C_(s))_(""^(n+3)C_(0)+""^(n+3)C_(1) -""^(n+3)C_(2))`
`=(-3!)/((n+1)(n+2)(n+3)){0-1+(n+3)-((n+3)(n+2))/(2!)}`
`=(-3!)/((n+1)(n+2)(n+3)).((n+2)(2-n-3))/(2) = (3!)/(2(n+3))`


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