1.

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

Answer» 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`
`n=(101-3)/2=98/2=49`
2)`(n+1)! =12(n-1)!`
`(n+1)n(n-1)! =12(n-1)!`
`n^2+n-12=0`
`(n+4)(n-3)=0`
`n=-4,3`
n can not be -4
n=3.


Discussion

No Comment Found

Related InterviewSolutions