Question

if (n+2)!=60[(n-1)!] find n​

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

Answer 2

$\begin{aligned}&(n+2)!=60(n-1)!\\ \\ \Longrightarrow\ \ &\frac{(n+2)!}{(n-1)!}=60\\ \\ \Longrightarrow\ \ &\frac{1\cdot 2\cdot 3\cdot...\cdot (n-1)n(n+1)(n+2)}{1\cdot 2\cdot 3\cdot ...\cdot(n-1)}=60\\ \\ \Longrightarrow\ \ &n(n+1)(n+2)=60\\ \\ \Longrightarrow\ \ &n^3+3n^2+2n=60\\ \\ \Longrightarrow\ \ &n^3+3n^2+2n-60=0\\ \\ \Longrightarrow\ \ &n^3-3n^2+6n^2-18n+20n-60=0\\ \\ \Longrightarrow\ \ &n^2(n-3)+6n(n-3)+20(n-3)=0\\ \\ \Longrightarrow\ \ &(n-3)(n^2+6n+20)=0\end{aligned}$

Since  $n^2+6n+20$  has no real roots, we may take,

$\huge\boxed{\mathbf{n=3}}$

($n^2+6n+20$  has no real roots because  $b^2-4ac=6^2-4\cdot 1\cdot 20=36-80=-44<0$)

Let's check!!!

$(n+2)!=(3+2)!=5!=120=60\cdot 2=60(2)!=60(3-1)!=60(n-1)!$

Hence checked!