Question

(n+2)! = 60 x (n-1)!​

Answer 1

Answer:

We have,

(n+2)!=60(n−1)

As we know n!=(n−1) n!

Therefore,

(n+2)(n+1)n×(n−1)!=60(n−1)!

(n+2)(n+1)n=60

(n

2

+2n+n+2)n=60

(n

3

+3n

2

+2n−60)=0

⇒n3+3n2+2n−60=0

Factoring the left hand side, this equation resolves to

(n−3)(n

2

+6n+20)=0(n−3)(n2+6n+20)=0

Since it can be shown that(n2+6n+20)=0 has (n

2

+6n+20)=0 no real solution,

n−3=0

n=3

Hence, the value of n is 3.

Answer 2

Answer:

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

${n}^{3} + 3 {n}^{2} + 2n = 60$

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