Question

solve it.................please​

Answer 1

(n+2)! = 60[(n-1)!], n=?

(n+2)! = 60* ( (n-1)! )

(n+2)! = (n+2) * (n+1) * (n) * (n-1)!  

{ (n+2) * (n+1) * (n) * (n-1)! } = 60* ( (n-1)! )

=> { (n+2) * (n+1) * (n) * (n-1)! } = 60*( (n-1)! )  

=> n3 + 3n2 +2n - 60 = 0

for n=3 we get 0 Substitute n = 3  

=>( n3 + 3n2 +2n - 60 ) / [n-3]

quotient= n2 +6n + 20  

=> (n-3) , (n2 +6n + 20)

n3 + 3n2 +2n - 60 = 0

=> (n-3) * (n2 +6n + 20) = 0  

=> (n-3) = 0 or (n2 +6n + 20) = 0

So n = 3