Christopher J. answered • 05/18/20
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Berkeley Grad Math Tutor (algebra to calculus)
Mary:
(n+3)! = (n+3)*(n+2)*(n+1)*n*(n-1)!
(n+3)!/(n-1)! = [(n+3)*(n+2)*(n+1)*n*(n-1)!]/ (n-1)!
Mary C.
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2 Answers By Expert Tutors
Mohit S. answered • 05/19/20
Tutor
New to Wyzant
Practice is the best way to success but for practice to need gaud
(n+3)!/(n-1)! First expand (n+3)! Like this (n+3)!=(n+3)* (n+2)*(n+1)*n*(n-1)!
Now
(n+3)!/(n-1)! = (n+3)* (n+2)*(n+1)*n*(n-1)!/(n-1)!
= (n+2)*(n+1)*n
= n(n^2+3n+2).
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