how do you solve (k + 2)!/(k -1)!(3!) - (k!)/(k-3)!(3)!?

Question

the answer is meant to be k^2. nothe the stated problem is only true for k greater than or equal to 4

Answer 1

OK (k+2)! / (k-1)! = (k+2)(k+1)*k
k!/(k-3)! = k*(k-1)*(k-2)

Factor out the 3! and we get
1/3! * [(k+2)(k+1)*k - k*(k-1)*(k-2)]
k/3! * [(k+2)(k+1) - (k-1)*(k-2)]
= k/3! * [k^2 + 3k + 2 - k^2 + 3k - 2]
= k/6 * 6k
= k^2

Answer 2

(k + 2)!/(k -1)!(3!) - (k!)/(k-3)!(3)!

=(k -1)! k.(k+1).(k+2)/(k -1)!6
-(k-3)!(k-2)(k-1)k/(k-3)!6


=k.(k+1).(k+2)/6 - (k-2)(k-1)k.6

you may continue