Question

if (n+2)! =60[(n-1)]find n
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Answer 1

$\tt\huge{Given}$

• if (n+2)! =60[(n-1)]find n

$\tt\huge{To\:\:Find\:\:\:n}$

(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

$\tt\huge{So\:\:\:n=3}$