Find `n ,` if `(n+2)! =2550xxn !` (ii) `(n+1)! =12xx(n-1)!`

Explanation

1)`(n+2)! =2550*n!`

`(n+2)(n+1)n! =2550(n!)`

`(n+2)(n+1)=2550`

`n^2+3n+2=2550`

`n^2+3n-2548=0`

`n=(-3pmsqrt(9+4(2548)))/2`

`n=(-3pmsqrt(10201))/2`

`n=(-3pm101)/2`

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