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

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Answer 1 · 2018-03-02T15:11:43

#n^2+3n+2#

We can rewrite the numerator as:

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

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

We can cancel #(n)!# and #(n)!# out:

#=((n+2) * (n+1) * 1)/1#

#=(n+2) * (n+1)#

#=n^2+3n+2#

Thus, solved