How do you simplify #((n+1)!)/((n-2)! )#?

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Answer 1 · 2018-03-02T03:00:38

#n^3-n#

The numerator can be rewritten as:

#((n+1) * (n+1-1) * (n+1-2) * (n+1-3)!)/((n-2)!)#

Simplifying:

#((n+1) * (n) * (n-1) * (n-2)!) / ((n-2)!)#

We can cancel the #(n-2)!# values out:

#(n+1)(n-1)(n)#

#=n^3-n#

And there we have our answer.