If `(sum_(n=1)^(n=n) (n^(2)+3n+3)(n+1)!)/((n+2)!)=8`, then n is equal to :

Answer

6

Explanation

`because" "underset(n=-1)overset(n)(Sigma)(n^(2)+3n+3)(n+1)! = 8(n+2)!`
`impliesunderset(n=-1)overset(n)(Sigma)((n+2)^(2)-(n+1)(n+1)! = 8 (n+2)!`
`impliesunderset(n=-1)overset(n)(Sigma)((n+2).(n+2)!-(n+1)(n+1)!)=8(n+2)!`
`implies(n+2).(n+2) ! = 8 (n+2)!`
`implies n+2=8impliesn=6`