Question

simplifique
a) (n-5)!/(n-3)
b) (2n+3)!/(2n+1)!

Answer 1

a) $\dfrac{(n-5)!}{(n-3).(n-4).(n-5)!} =\dfrac{1}{(n-3).(n-4)}=\dfrac{1}{n^{2}-7n+12 }$

b) $\dfrac{(2n+3)!}{(2n+1)!} =\dfrac{(2n+3).(2n+2).(2n+1)!}{(2n+1)!} =(2n+3).(2n+2)$$=4 n^{2} +10n+6$