How do you simplify #(3(n+1)! )/ (5n!)#?

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Answer 1 · 2015-11-09T15:35:35

#{3(n+1)!}/{5n!} = {3(n+1)}/5#

The factorial of any integer #n# is given by

#n! = n(n-1)(n-2)(n-3)...(3)(2)(1)#

Therefore, the factorial of #n+1# would be

#(n+1)! = (n+1)(n)(n-1)(n-2)(n-3)...(3)(2)(1)#

Using these two expression in our original question,

#{3(n+1)!}/{5n!} = [3(n+1)(n)(n-1)(n-2)...(3)(2)(1)]/[5(n)(n-1)(n-2)...(3)(2)(1)]#

All terms in the denominator except #5# are cancelled out by corresponding terms in the numerator. Therefore, you finally get

#{3(n+1)!}/{5n!} = {3(n+1)}/5#